Foundations of Point Set Topology
Frederic Latremoliere
August 17, 2011

1. Topological Spaces and continuous functions

1.1. The category of Topological spaces

Definition 1.1.

Let E be a set. A topology 𝒯 on E is a subset of 𝒫E such that:

(Top 1)

E𝒯 and 𝒯.

(Top 2)

For all U,V𝒯 one has UV𝒯.

(Top 3)

For all 𝒰𝒯 one has 𝒰𝒯.

A pair E,𝒯 is a topological space when E is a set and 𝒯 is a topology on E. Moreover, a subset U of E is called open, if U𝒯E, and closed, if EU𝒯E.

The section on fundamental examples collects many examples of topologies. Note that ,E and 𝒫E both are obvious topologies on any set E, but of course the examples in the next section are more substantial. For now, we will work out a few basic properties of topologies.

Proposition 1.2.

Let E be a (nonempty) set, F,𝒯F be a topological space, and f:EF a function. Define:

𝒯f=f-1VV𝒯F.

Then E,𝒯f is a topological space. One calls 𝒯f the initial topology for f or the topology induced by f.

Proof.

By construction, f-1F=E and f-1= so ,E𝒯f. Let U1,U2𝒯f. Then by definition, there exists V1,V2𝒯F such that Ui=f-1Vi for i=1,2. Thus:

U1U2=f-1V1f-1V2=f-1V1V2.

Hence by definition, U1U2𝒯F. Last, let 𝒰𝒯f. By definition, each element of 𝒰 is the preimage by f of some open set VU in F,𝒯F. We set:

V=V𝒯F:U𝒰U=f-1V.

Now x𝒰 if and only if there exists U𝒰 such that xU. Now, U𝒰 if and only if there exists VV such that U=f-1V. Hence x𝒰 if and only if there exists VV such that xf-1V, which in turn is equivalent to xf-1V. Hence:

𝒰=f-1V

and therefore, we have shown that 𝒯f is a topology on E. ∎

Definition 1.3.

Let E,𝒯E and F,𝒯F be two topological spaces. Let f:EF be a function. We say that f is continuous on E when:

V𝒯Ff-1V𝒯E.
Remark 1.4.

It is immediate that f:EF is continuous if and only if for all closed subset CF, the set f-1C is closed as well.

Remark 1.5.

It is immediate by definition that any constant function is continuous. It is also obvious that the identity function on any topological space is continuous.

Proposition 1.6.

Let E,𝒯E, F,𝒯F and G,𝒯G be three topological spaces. Let f:EF and g:FG. If f and g are continuous, so is gf.

Proof.

Let V𝒯G. Then g-1V𝒯F by continuity of g. Hence f-1g-1V𝒯E by continuity of f. ∎

Definition 1.7.

The category whose objects are topological spaces and morphisms are continuous functions is the category of topological spaces.

Definition 1.8.

An isomorphism in the category of topological spaces is called an homeomorphism.

Remark 1.9.

An homeomorphism is by definition a continuous map which is also a bijection and whose inverse function is also continuous. There are continuous bijections which are not homeomorphisms (see the chapter Fundamental Examples).

1.2. Order on Topologies on a given set

The initial topology 𝒯f induced by a function f:EF is a subset of the topology on E if and only if f is continuous. This motivates the following definition.

Definition 1.10.

Let E be a set. Let 𝒯1 and 𝒯2 be two topologies on E. We say that 𝒯1 is finer than 𝒯2 and 𝒯2 is coarser than 𝒯1 when 𝒯2𝒯1.

Of course, inclusion induces an order relation on topologies on a given set. A remarkable property is that any nonempty subset of the ordered set of topologies on a given set always admits a greatest lower bound.

Theorem 1.11.

Let E be a set. Let 𝒯 be a nonempty set of topologies on E. Then the set:

𝒯=𝒯=UE:σ𝒯Uσ

is a topology on E and it is the greatest lower bound of 𝒯 (where the order between topology is given by inclusion).

Proof.

We first show that 𝒯 is a topology. By definition, for all σ𝒯, since σ is a topology on E, we have E,σ. Hence ,E𝒯. Now, let U,V𝒯. Let σ𝒯 be arbitrary. By definition of 𝒯, we have Uσ and Vσ. Therefore, UVσ since σ is a topology. Since σ was arbitrary in 𝒯, we conclude that UV𝒯 by definition. The same proof can be made for unions. Let 𝒰𝒯. Let σ𝒯 be arbitrary. By definition of 𝒯, we have 𝒰σ. Since σ is a topology, 𝒰σ. Hence, as σ was arbitrary, 𝒰𝒯. So 𝒯 is a topology on E. By construction, 𝒯σ for all σ𝒯, so 𝒯 is a lower bound for 𝒯. Assume given a new topology ρ on E such that ρσ for all σ in 𝒯. Let Uρ. Then for all σ𝒯 we have Uσ. Hence by definition U𝒯. So ρ𝒯 and thus 𝒯 is the greatest lower bound of 𝒯. ∎

Corollary 1.12.

Let E be a set and F,𝒯F be a topological space. The smallest topology which makes a function f:EF continuous is 𝒯f.

Proof.

Let 𝒯 be the set of all topologies on E such that f is continuous. By definition, 𝒯f is a lower bound of 𝒯. Moreover, 𝒯f𝒯. Hence it is the greatest lower bound. ∎

We can use Theorem (1.11) to define other interesting topologies. Note that trivially 𝒫E is a topology, so given any A𝒫E for some set E, there is at least one topology contaning A. From this:

Definition 1.13.

Let E be a set. Let A be a subset of 𝒫E. The greatest lower bound of the set:

𝒯A=σis a topology on E:Aσ

is the smallest topology on E containing A. We call it the topology induced by A on E and we denote it 𝒯A.

We note that since 𝒯A belongs to 𝒯A by construction, it is indeed the smallest topology containing A.

In general, we will want a better description of the topology induced by a particular set than the general intersection above. This is not always possible, but the concept of basis allows one to obtain useful descriptions of topologies.

1.3. Basis

When inducing a topology from a family of subsets of some set E, the fact that enjoys the following property greatly simplifies our understanding of 𝒯.

Definition 1.14.

Let E be a set. A topological basis on E is a subset of 2E such that:

  1. E=,

  2. For all B,C such that BC, we have:

    xBCDxDDBC.

The main purpose for this definition stems from the following theorem:

Theorem 1.15.

Let E be some set. Let be a topological basis on E. Then the topology induced by is:

𝒯=𝒰:𝒰.
Proof.

Denote, for this proof, the set 𝒰:𝒰, by σ, and let us abbreviate 𝒯 by 𝒯. We wish to prove that 𝒯=σ. First, note that σ by construction. By definition, 𝒯 and since 𝒯 is a topology, it is closed under arbitrary unions. Hence σ𝒯. To prove the converse, it is sufficient to show that σ is a topology. As it contains and 𝒯 is the smallest such topology, this will provide us with the inverse inclusion. By definition, = and thus σ. By assumption, since is a basis, E= so Eσ. As the union of unions of elements in is a union of elements in , σ is closed under abritrary unions. Now, let U,V be elements of . If UV= then UVσ. Assume that U and V are not disjoints. Then by definition of a basis, for all xUV there exists Wx such that xWx and WxUV. So:

UV=xUVWx

and therefore, by definition, UVσ. We conclude that the intersection of two arbitary elements in σ is again in σ by using the distributivity of the union with respect to the intersection. ∎

The typical usage of this theorem is the following corollary. We shall say that a basis on a set E is a basis for a topology 𝒯 on E when the smallest topology containing is 𝒯.

Corollary 1.16.

Let be a topological basis for a topology 𝒯 on E. A subset U of E is in 𝒯 if and only if for any xU there exists B such that xB and BU.

Proof.

We showed that any open set for the topology 𝒯 is a union of elements in : hence if xU for U𝒯 then there exists B such that xB and BU. Conversely, if U is some subset of E such that for all xU there exists Bx such that xBx and BxU then U=xUBx and thus U𝒯. ∎

As a basic application, we show that:

Corollary 1.17.

Let E,𝒯E and F,𝒯F be two topological spaces. Let be a basis for the topology 𝒯E. Let f:EF. Then f is continuous on E if and only if:

V𝒯Fxf-1VBxBBf-1V.
Corollary 1.18.

Let E,𝒯E and F,𝒯F be two topological spaces. Let be a basis for the topology 𝒯F. Let f:EF. Then f is continuous on E if and only if:

Vf-1V𝒯E.
Proof.

By definition, continuity of f implies 1.18. Conversely, assume 1.18 holds. Let V𝒯F. Then there exists 𝒰 such that V=𝒰. Now by assumption, f-1B𝒯E for all B𝒰 and thus f-1V=B𝒰f-1B𝒯E since 𝒯E is a topology. ∎

We leave to the reader to write the statement when both E and F have a basis.

1.4. Interior and Closure

Given a topological space E,𝒯, an arbitrary subset A of E may be neither open nor closed. It is useful to find closest “approximations” for A which are either open or closed. The proper notions are as follows.

Proposition 1.19.

Let E,𝒯 be a topological space. Let AE. There exists a (necessarily unique) largest open set in E contained in A and a (necessarily unique) smallest closed set in E containing A (where the order is given by inclusion).

Proof.

Let 𝒰=U𝒯:UA. Then since 𝒯 is a topology, 𝒰𝒯. Moreover, by definition, if x𝒰 then there exists U𝒰 such that xU and since UA we conclude xA. Therefore 𝒰 is by construction the largest open subset of E contained in A.

The reasonning is similar for the smallest closed set. Namely, let =FE:EF𝒯AF. Since closed sets are the complements of open sets, it follows that arbitrary intersections of closed sets is closed. Thus is a closed set, and it is the smallest containing A by construction. ∎

Definition 1.20.

Let E,𝒯 be a topological space. Let AE. The interior of A is the largest open subset of E contained in A denoted by Å. The closure of A is the smallest closed subset of E containing A and is denoted by A¯.

Thus we always have ÅAA¯. Moreover:

Proposition 1.21.

Let E,𝒯 be a topological space. Let A,BE. Then:

  1. If AB then ÅB̊ and A¯B¯,

  2. AB̊=ÅB̊,

  3. AB¯=A¯B¯.

Proof.

If AB then ÅAB. Since Å is open and a subset of B, it is contained in B̊ by definition. Similarly, if AB then AB¯. Now, B¯ is closed and contains A so it must contained the smallest closed subset containing A, namely A¯.

Note that ABA so AB̊Å. Similarly with B, so AB̊ÅB̊. On the other hand, ÅB̊ is an open set (as the intersection of two open sets), and since ÅA and B̊B, it is included in AB. Hence it is included in the largest open subset contained in AB, which completes this proof.

The same reasonning can be applied to the last assertion of this proposition. ∎

The main theorem about closures is the following.

Theorem 1.22.

Let E,𝒯 be a topological space. Let AE. Then xA¯ if and only if for all V𝒯 such that xV we have VA.

Proof.

Assume first that xA¯. Then xEA¯ and EA¯ is open by definition; moreover by construction EA¯A= since AA¯ so EA¯EA. Conversely, let xE and assume that there exists V𝒯 such that xV and yet VA=. Then AEV and EV is closed, so by definition A¯EV, hence xA¯. ∎

In general, one can thus write:

A¯=xE:V𝒯xVVA.

The following is immediate, so we omit the proof.

Corollary 1.23.

Let E,𝒯 be a topological space and a basis for 𝒯. Let AE. Then xA¯ if and only if for all V such that xV we have VA.

We can use the main theorem about closure to show that:

Proposition 1.24.

Let E,𝒯 be a topological space. Let AE. Then EA¯=EÅ and EÅ=EA¯.

Proof.

We shall only prove that EA¯=EÅ as both assertions are proved similarly. Note first that ÅA so EAEÅ and EÅ is closed. Hence EA¯EÅ. Conversely, let xEÅ. Let V𝒯 such that xV. Assume VEA=. Then VA so, as V open, VÅ, which contradicts xEÅ. So VEA, and thus xEA¯ as desired. ∎

1.5. Characterizations of Continuity

Theorem 1.25.

Let E,𝒯E and F,𝒯F be topological spaces. Let f:EF be given. Then f is continuous if and only if for all AE, fA¯fA¯.

Proof.

Assume first that f is continuous. Let AE. Since f is continuous, f-1fA¯ is closed, and by definition of images and preimages,

Af-1fA¯

so A¯f-1fA¯ as desired.

Let us now assume that for any AE, we have fA¯fA¯. Then let CF be a closed set. Let D=f-1C. By assumption, D¯f-1fD¯f-1fD¯f-1C¯=f-1C=D. So D¯D, hence D=D¯ so D is closed. Hence f is continous. ∎

Remark 1.26.

In the next chapter on fundamental examples, one will encounter many topological spaces where singletons are not open. Note that in those cases, if f:eF is constant, then fE̊ would be empty, and thus we can not expect fE̊fE̊ in general for continuous functions.

Theorem 1.27.

Let E,𝒯E and F,𝒯F be topological spaces. Let f:EF be given. Then f is continuous if and only if for all AF, we have f-1Åf-1Å.

Proof.

Assume first that f is continuous. Let AF. Then f-1Å𝒯E since f is continuous. Since f-1Åf-1A, we conclude that f-1Åf-1Å (by maximality of f-1Å among all open sets in E contained in A).

Conversely, assume that for all AF, f-1Åf-1Å. Let V𝒯F. Then by assumption:

f-1V=f-1V̊f-1V̊f-1V

so f-1V=f-1V̊ i.e. f-1V𝒯E. Since V𝒯F was arbitrary, f is continuous. ∎

Theorem 1.28.

Let E,𝒯E and F,𝒯F be topological spaces. Let f:EF be given. Then f is continuous if and only if for all AF we have f-1A¯f-1A¯.

Proof.

Assume first that f is continuous. Let AF. Then AA¯ so f-1Af-1A¯. Since f is continuous, then f-1A¯ is closed, and thus by minimality of f-1A¯, we have f-1A¯f-1A¯.

Conversely, assume that for all AF we have f-1A¯f-1A¯. Let CF be closed. Then:

f-1C¯f-1C¯=f-1C

so f-1C=f-1C¯ i.e. f-1C is closed. Since C was an abritrary closed subset of F, the function f is continuous. ∎

1.6. Topologies defined by functions

Continuity was phrased by stating the topology induced by the function is coarser than the topology on the domain. We can extend this idea to define topologies. Note that if E is endowed with the indiscrete topology 2E then any function, to any topological space, is continuous. Hence the following definition can be made:

Definition 1.29.

Let E be a set. Let F,𝒯F be a topological space. Let be a nonempty set of functions from E to F. The smallest topology on E such that all the functions in are continuous is the initial topology induced by on E. We denote it by 𝒯.

Proposition 1.30.

Let E be a set. Let F,𝒯 be a topological space. Let be a nonempty set of functions from E to F. Let:

=Bf1,U1,,fn,Un:f1,,fn,U1,,Un𝒯F

where

Bf1,U1,,fn,Un=f1-1U1fn-1Un

for all n-tuple f1,,fn of functions in and U1,,Un𝒯F. Then is a basis for the initial topology induced by .

Proof.

Note that for all f, we must have 𝒯f𝒯 by defintion of continuity. Hence, if 𝒯 is any topology on E such that all the functions in are continuous, then 𝒯. In particular, 𝒯𝒯. On the other hand, by construction, every function in is continuous for 𝒯, so 𝒯𝒯 by definition of the initial topology. Hence 𝒯=𝒯.

It remains to show that is a basis on E. By definition, note that E since E=f-1F for any f. Now, note that by definition, the intersection of two elements in is still in , so is a trivially a basis. ∎

The main two theorems regarding initial topologies describe its universal property:

Theorem 1.31.

Let E be a set. Let F,𝒯F and D,𝒯D be topological spaces. Let be a nonempty set of functions from E to F. A function f:DE is continuous when E is endowed with the initial topology 𝒯 if and only if gf is continuous for all g.

Proof.

If f is continuous, then for all g the function gf is continuous, as the composition of two continuous functions.

Conversely, assume that gf is continuous for all g. Let g1,,gn and V1,,Vn𝒯F and set:

W=g1-1V1gn-1Vn.

Now, since g1f,,gnf are all continuous by assumption, f-1g1-1V1,,f-1gnVn are all in 𝒯D. Hence f-1W𝒯D. Since W was an arbitrary set in a basis for the final topology on E, f is continuous. ∎

Theorem 1.32.

Let E be a set. Let F,𝒯F and D,𝒯D be topological spaces. Let be a nonempty set of functions from E to F. The initial topology for is the unique topology such that, given any topological space D,𝒯D and given any function f:DE, then f is continuous if and only if gf is continuous for all g.

Proof.

We showed that the initial topology for has the desired property. Conversely, assume that the given property holds for some topology 𝒯 on E. In particular, let f:EE be the identity, seen as a function from the topological space E,𝒯 into the space E,𝒯. Using the specified universal property, we see that f is continuous, so 𝒯 is finer than 𝒯. Conversely, note that f-1 is also continuous for the same reason, so 𝒯 is finer than 𝒯, so these two topologies agree. This completes the proof of this universal property. ∎

Last, we can obtain a basis for initial topologies in a natural way:

Theorem 1.33.

Let E be a set, F,𝒯F be a topological space, a nonempty set of functions from E to F. Let be a basis for 𝒯F. Then the set:

𝒞=f1-1B1fn-1Bn:f1,,fn,B1,,Bn

is a basis for the initial topology for .

Proof.

This is a quick computation. ∎

The dual notion of initial topology is final topology:

Definition 1.34.

Let F be a set. Let be a set of triplets E,𝒯,f where E,𝒯 is a topological space, and f:EF is a function. The final topology 𝒯 on F is the smallest topology such that all fuctions f such that E,𝒯,f are continuous.

In the following chapters, we will see that the product topology, the trace topology and the norm topology are examples of initial topologies. The quotient topology is an example of a final topology.

2. Fundamental Examples

This chapter provides various examples of topological spaces which will be used all along these notes and are often at the core of the subject.

2.1. Trivial Topologies

Definition 2.1.

Let E be a set. The topology ,E is the indiscrete topology on E. The topology 2E is the discrete topology on E.

Proposition 2.2.

Let F,𝒯 be a topological space. All functions from E,2E into F,𝒯 are continuous. The only continuous functions from E,,E to F are constant if F,𝒯 is T1.

Proof.

When E is given the discrete topology, then for all open subsets V of F one has f-1V open in E. On the other hand, if E is given the indiscrete topology and F,𝒯 is T1 then assume f takes two values, l and k. Then Fl is open, so f-1Fl must be open, and as it is nonempty (it contains f-1k), it is all of E, which is absurd. ∎

2.2. Order topologies

Definition 2.3.

Let (E,) be a linearly ordered set. Let ,- be two symbols not in E. Define E¯=E-, and extend to E¯ by setting for all x,yE¯:

xy(xEyExy)(x=-)(y=).

Let < be the relation (i.e. x<y(xyxy)) on E¯. Let a,bE¯. We define the set a,b=xE:a<xx<b.

Definition 2.4.

Let (E,) be a linearly ordered set. Then the topology induced by the set:

IE=a,b:a,bE¯

is the order topology on E.

Proposition 2.5.

The set IE is a basis for the order topology on E.

Proof.

Since E is linearly ordered, so is E¯. It is immediate that a,bc,d=a,d if a is the largest of a and c and d is the smallest of b and d. ∎

Remark 2.6.

The default topology on is the order topology.

Definition 2.7.

Let (E,) be an ordered set, and a,bE. We define a,b=xE:axxb.

2.3. Trace Topologies

Proposition 2.8.

Let E,𝒯E be a topological space. Let AE. Then the trace topology on A induced by 𝒯E is the topology:

𝒯AE=UA:U𝒯E.
Proof.

It is a trivial exercise to show that the above definition indeed gives a topology on A. ∎

Just as easy is the following observation:

Proposition 2.9.

Let E,𝒯 be a topological space. Let be a basis for 𝒯. Let AE. Then the set BA:B is a basis for the trace topology on A induced by 𝒯.

Proof.

Trivial exercise. ∎

Remark 2.10.

The default topology on , and is the trace topology induced by the order topology on . Since n=n-12,n+12 for all n, we see that the natural topologies form and are in fact the discrete topology.

Remark 2.11.

Let AE, where E,𝒯 is a topological space. Let i:AE be the inclusion map. Then the trace topology is the initial topology for i, i.e. 𝒯i.

2.4. Product Topologies

Definition 2.12.

Let I be some nonempty set. Let us assume given a family Ei,𝒯iiI of topological spaces. A basic open set of the cartesian product iIEi is a set of the form iIUi where iI:UiEi is finite and for all iI, we have Ui𝒯i.

Definition 2.13.

Let I be some nonempty set. Let us assume given a family Ei,𝒯iiI of topological spaces. The product topology on iIEi is the smallest topology containing all the basic open sets.

Proposition 2.14.

Let I be some nonempty set. Let us assume given a family Ei,𝒯iiI of topological spaces. The collection of all basic open sets is a basis on the set iIEi.

Proof.

Trivial exercise. ∎

Remark 2.15.

The product topology is not just the basic open sets on the cartesian products: there are many more open sets!

Proposition 2.16.

Let I be some nonempty set. Let us assume given a family Ei,𝒯iiI of topological spaces. The product topology on iIEi is the initial topology for the the set pi:iI where pi:jIEjEi is the canonical surjection for all iI.

Proof.

Fix iI. Let V𝒯Ei. By definition, pi-1V=jIUj where Uj=Ej for jIi, and Ui=V. Hence pi-1V is open in the product topology. As V was an arbitrary open subset of Ei, the map pi is continuous by definition. Hence, as i was arbitrary in I, the initial topology for pi:iI is coarser than the product topology.

Conversely, note that the product topology is generated by pi-1V:iI,V𝒯Ei, so it is coarser than the initial topology for pi:iI. This concludes this proof. ∎

Corollary 2.17.

Let I be some nonempty set. Let us assume given a family Ei,𝒯iiI of topological spaces. Let 𝒯 be the product topology on F=iIEi. Let D,𝒯D be a topological space. Then f:DF is continuous if and only if pif is continuous from D,𝒯D to Ei,𝒯Ei for all iI, where pi is the canonical surjection on Ei for all iI.

Proof.

We simply applied the fundamental property of initial topologies. ∎

Remark 2.18.

The box topology on the cartesian product is the smallest topology containing all possible cartesian products of open sets. It is finer than the product topology in general. Since the product topology is the coarsest topology which makes the canonical projections continuous, it is the preferred one on cartesian products. Of course, both agree on finite products.

Remark 2.19.

The product topology is the default topology on a cartesian product of topological spaces.

2.5. Metric spaces

Definition 2.20.

Let E be a set. A function d:E × E0, is a distance on E when:

  1. For all x,yE, we have dx,y=0 if and only if x=y,

  2. For all x,yE we have dx,y=dy,x,

  3. For all x,y,zE we have dx,ydx,z+dz,y.

Definition 2.21.

A pair E,d is a metric space when E is a set and d a distance on E.

The following is often useful:

Proposition 2.22.

Let (E,d) be a metric space. Let x,y,zE. Then:

dx,y-dx,zdy,z.
Proof.

Since dx,ydx,z+dz,y we have dx,y-dx,zdz,y=dy,z. Since dx,zdx,y+dy,z we have dx,z-dx,ydy,z. Hence the proposition holds. ∎

Definition 2.23.

Let E,d be a metric space. Let xE and r0,. The open ball of center x and radius r in E,d is the set:

Bx,r=yE:dx,y<r.
Definition 2.24.

Let E,d be a metric space. The metric topology on E induced by d is the smallest topology containing all the open balls of E.

Theorem 2.25.

Let E,d be a metric space. The set of all open balls on E is a basis for the metric topology on E induced by d.

Proof.

It is enough to show that the set of all open balls is a basis. By definition, E=xEBx,1. Now, let us be given Bx,rx and By,ry for some x,yE and rx,ry>0. If the intersection of these two balls is empty, we are done; let us assume that there exists zBx,rxBy,ry. Let ρ be the smallest of rx-dx,z and ry-dy,z. Let wBz,ρ. Then:

dx,wdx,z+dz,w<dx,z+rx-dx,z=rx

so wBx,rx. Similarly, wBy,ry. Hence, Bz,ρBx,rxBy,ry as desired. ∎

The following theorem shows that metric topologies are minimal in the sense of making the distance functions continuous.

Theorem 2.26.

Let E,d be a metric space. For all xE, the function yEdx,y is continuous on E for the metric topology. Moreover, the metric topology is the smallest topology such that all the functions in the set ydx,y:xE are continuous.

Proof.

Fix xE. It is sufficient to show that the preimage of 0,r and r, by dx:yEdx,y is open in the metric topology of E, where r0 is arbitrary. Indeed, these intervals form a basis for the topology of 0,. Let r0 be given. Then dx-10,r=Bx,r by definition, so it is open. Moreover, it shows that the minimal topology making all these maps continuous must indeed contain the metric topology. Now, let yE such that dx,y>r. Let ρ=dx,y-r>0. Then if dw,y<ρ for some wE then:

dx,ydx,w+dw,yso dx,y-dw,ydx,w

so dx,w>r. Hence

By,ρdx-1r,

for all ydx-1r,. Therefore, dx-1r, is open, as desired, and our proposition is proven. ∎

Remark 2.27.

The topology on 0, is the trace topology on 0, induced by the usual, i.e. the order topology on .

Remark 2.28.

The metric topology is the default topology on a metric space.

There are more examples of continuous functions between metric spaces. More precisely, a natural category for metric spaces consists of metric spaces and Lipschitz maps as arrows, defined as follows:

Definition 2.29.

Let E,dE, F,dF be metric spaces. A function f:EF is k-Lipschitz for k0, if:

x,yEdFfx,fykdEx,y.
Definition 2.30.

Let E,dE, F,dF be metric spaces. Let f:EF be a Lipschitz function. Then the Lipschitz constant of f is defined by:

Lipf=supdFfx,fydEx,y:x,yE,xy.
Remark 2.31.

Lipf=0 if and only if f is constant.

Proposition 2.32.

Let E,dE, F,dF be metric spaces. If f:EF is a Lipschitz function, then it is continuous.

Proof.

Assume f is nonconstant (otherwise the result is trivial). Let k be the Lipschitz constant for f. Let yF and ϵ>0. Let xf-1By,ϵ. Let zE such that dEx,z<δx=ϵ-dfx,yk (note that the upper bound is nonzero).

(2.1) dFfz,ydFfz,fx+dFfx,y
(2.2) kdEx,z+dFfx,y
(2.3) <ϵ-dFfx,y+dFfx,y=ϵ.

Hence f-1By,ϵ=xf-1By,ϵBx,δx. So f is continuous. ∎

Remark 2.33.

The proof of continuity for Lipshitz maps can be simplified: it is a consequence of the squeeze theorem. We refer to the chapter on metric spaces for this.

Remark 2.34.

Using Lipshitz maps as morphisms for a category of metric spaces is natural. Another, more general type of morphisms, would be uniform continuous maps, which are discussed in the compact space chapter.

2.6. Co-Finite Topologies

A potential source for counter-examples, the family of cofinite topologies is easily defined:

Proposition 2.35.

Let E be a set. Let:

𝒯cofE=UE:EUis finite .

Then 𝒯cofE is a topology on E.

Proof.

By definition, 𝒯cofE. Moreover, EE= which is finite, so E𝒯cofE. Let U,V𝒯cofE. If U or V is empty then UV= so UV𝒯cofE. Otherwise, EUV=EUV which is finite, since by definition EU and EV are finite. Hence UV𝒯cofE. Last, let 𝒰𝒯cofE. Again, if 𝒰= then 𝒰=𝒯cofE. Let us now assume that 𝒰 contains at least one nonempty set V. Then:

E𝒰=EU:U𝒰EV.

Since EV is finite by definition, so is 𝒰, which is therefore in 𝒯cofE. This completes our proof. ∎

2.7. The one-point compactification of

Limits of sequences is a central tool in topology and this section introduces the natural topology for this concept. The general notion of limit is the subject of the next chapter.

Definition 2.36.

Let be some symbol not found in . We define ¯ to be .

Proposition 2.37.

The set:

𝒯¯={U¯:(U)(UUis finite)}

is a topology on ¯.

Proof.

By definition, so 𝒯¯. Moreover ¯¯= which has cardinal 0 so ¯𝒯¯. Let U,V𝒯¯. If either U or V is a subset of then UV is a subset of so UV𝒯¯. Othwiwse, UV. Yet ¯UV=¯U¯V which is finite as a finite union of finite sets. Hence UV𝒯¯ again.

Last, assume that 𝒰𝒯¯. Of course, 𝒰 if and only if U for some U𝒰. So, if 𝒰 then 𝒰𝒯¯ by definition. If, on the other hand, 𝒰, then there exists U𝒰 with ¯U finite. Now, ¯𝒰=¯V:V𝒰¯U so it is finite, and thus again 𝒰𝒯¯. ∎

3. Limits

3.1. Topological Separation and Hausdorff spaces

The general definition of topology allows for examples where elements of a topological space, seen as a set, can not be distinguished from each others (for instance if the topology is indiscrete). When points can be topologically differentiated, a topology is in some sense separated. There are many axioms, or definitions, of separability, and we will use the most common and intuitive: namely, the notion of Hausdorff spaces. We do however present a few basic notions in this section which are weaker than Hausdorff separation, as such spaces are certainly common in mathematics.

Definition 3.1.

Let E,𝒯 be a topological space. We say that 𝒯 is T0 when given any two points x,yE with xy, there exists an open set U𝒯 such that either xU,yU or yU,xU.

Thus a space is T0 when there are enough open sets to separate the points, i.e. when the set of all open sets containing one point is not the same as the set of all open sets containing a different point. However, this notion is not symmetric. The following definition add that the separation property should be symmetric:

Definition 3.2.

Let E,𝒯 be a topological space. We say that 𝒯 is T1 when given any two points x,yE with xy, there exists two open sets U,V𝒯 such that xU,yU and yV,xV.

A key advantage of T1 separation is:

Proposition 3.3.

Let E,𝒯 be a topological space. Then 𝒯 is T1 if and only if for all xE the set x is closed.

Proof.

Assume 𝒯 is T1. Fix xE. Let yEx. Then there exists Uy𝒯 such that xUy and yUy. We can thus write:

Ex=yExUy

which shows that Ex is open as desired.

Assume now that all singletons are closed. Let x,yE with xy. Then xEy and yEx, i.e. 𝒯 is T1. ∎

Note that T0 is not enough for the above result.

Corollary 3.4.

Let E,𝒯 be a topological space. E,𝒯 is T1 if and only if for all xE we have x=U𝒯:xU.

Proof.

Assume E,𝒯 is T1. Let S=U𝒯:xU. Let yS. If yx then there exists U𝒯 such that xU and yU. This is a contradiction. So S=x.

Assume now that for all xE we have x=U𝒯:xU. Let x,yE such that xy. Then:

U𝒯:xUU𝒯:yU

, so there exists U𝒯 such that xU and yU. Similarly:

U𝒯:yUU𝒯:xU

so there exists U𝒯 such that yU and xU. Hence E,𝒯 is T1. ∎

Example 3.5.

The indiscrete topology is not T0.

Example 3.6.

Let E be an infinite set (for instance E=), endowed with the cofinite topology 𝒯=𝒯cofE. By defintion, x is closed for all xE so E,𝒯 is T1. We make a useful observation. Let f:E be a continuous map, where the codomain is endowed with the usual order topology. Assume f is not constant: then there exists x,yE such that fxfy. Without loss of generality, we assume fx<fy. Let r=12fy-fx. Let U=fx-r,fx+r and V=fy-r,fy+r. Then UV= and U,V are open sets in such that xU and yV. Since f is continuous, f-1U and f-1V are open in E, i.e. are cofinite. Since U and V are disjoint, we conclude that f-1U is in the complement of f-1V, which is finite. Since the complement of f-1U is finite as well, we conclude that E is finite, which is a contradiction. So f is a constant.

The cofinite topology on infinite set example shows that T1 still allows for counter intuitive situations. We also saw that we could find two disjoint open sets in containing given distinct points: this stronger property is the separation axiom we will focus our attention to.

Definition 3.7.

Let E,𝒯 be a topological space. We say that E,𝒯 is a Hausdorff space (or T2) when for any x,yE such that xy there exists U,V𝒯 such that UV=, xU and yV.

Proposition 3.8.

If E,𝒯 is Hausdorff, then it is T1.

Proof.

This result holds by definition. ∎

Example 3.9.

Let (E,) be a linearly ordered set and let 𝒯 be the associated order topology on E. Then E,𝒯 is Hausdorff. Indeed, let x,yE with xy; without loss of generality we assume that x<y. Then if there exists zE such that x<z<y then x-,z and yz,, where both intervals are disjoint and open. Otherwise, -,yx,= and x-,y,yx,.

Example 3.10.

In particular, is Hausdorff.

Example 3.11.

Let E,𝒯 be a Hausdorff space. Then by definition, it is immediate that the trace topology induced on any subset of E is also Hausdorff.

Example 3.12.

In particular, the topology on is Hausdorff.

Example 3.13.

The discrete topology on any set is always Hausdorff.

Example 3.14.

In particular, the topology on and is Hausdorff.

Example 3.15.

Let E,d be a metric space. Then E,𝒯d is Hausdorff. Indeed, let x,yE such that xy. Let r=12dx,y and note that r>0 by definition of a distance. Let zBx,r. Then:

dx,ydx,z+dz,ysor<dz,y

and by symmetry, if dw,y<r then dw,x>r. Hence Bx,rBy,r=.

Example 3.16.

Let Ei,𝒯iiI be some family of Hausdorff topological spaces. Then the cartesian product with the product topology is Hausdorff. Indeed, let x=xiiI and y=yiiI such that xy. By definition, there exists i0I such that xi0yi0. Since Ei0 is Hausdorff, there exists Ui0,Vi0 such that xi0Ui0, yi0Vi0 and Ui0Vi0=. Define the two families:

u:iIEiif ii0,Ui0otherwise,

and

v:iIEiif ii0,Vi0otherwise,

and set U=iIui and V=iIvi. Then by construction, xU, yV and UV=.

The following result is a characterization of Hausdorff separation.

Theorem 3.17.

Let E,𝒯 be a topological space. Then E,𝒯 is Hausdorff if and only if Δ=x,x:xE is closed in E2 for the product topology.

Proof.

Assume E is Hausdorff. Let x,yE2Δ (so xy!). Then there exists two disjoint open sets U and V in E such that xU and yV. If z,zU × V then zU and zV which is impossible since U and V are disjoint. So U × V, which is open in the product topology, contains x,y by definition, and is a subset of E2Δ. Hence E2Δ is open.

Conversely, assume E2Δ is open. Let x,yE2Δ. Since basic open sets form a basis for the product topology, there exists U,V open in E such that x,yU × V and U × VE2Δ. Now, by definition, U × VΔ= so as above, UV=, as desired. So E,𝒯 is Hausdorff. ∎

Hausdorff spaces have a nice relation with continuous maps as well.

Proposition 3.18.

Let E,𝒯E be a topological space. Let F,𝒯F be a Hausdorff topological space. Let f:EF and g:EF be given.

  1. If f is continuous, then the kernel kerf=x,yE2:fx=fy of f is closed.

  2. If f is continuous, then the graph graph(f)={(x,f(x))E × F} of f is closed.

  3. If f and g are continuous, then the equalizer eqf,g=xE:fx=gx of f,g is closed.

Proof.

Since F,𝒯F is Hausdorff, the set Δ=x,x:xF is closed.

We first prove that the kernel of f is closed when f is continuous. Let:

κ:E × EF × F
x,yfx,fy.

The function x,yE × Ex is continuous since the product topology is the initial topology for the canonical surjections, and f is continuous by assumption, so x,yE × Efx is continuous as the composition of two continuous functions. Similarly, x,yE × Efy is continuous. Hence, by the universal property of the initial topology, κ is continuous.

Now, by definition, kernelf=κ-1Δ and thus it is closed.

We now prove that the graph of f is closed when f is continuous. Let:

δ:E × FF × F
x,yfx,y.

By a similar argument as for κ, the map δ is continuous since x,yE × Fy and x,yE × Fx are continuous for the product topology, and f is continuous, so by composition of continuous functions, x,yE × Ffx is continuous. Hence by the universal property of the product topology (seen again as the initial topology for the canonical surjections), δ is continuous.

Now graphf=δ-1Δ so the graph of f is closed.

Last, we show the equalizer of f and g is closed when both f and g are continuous. Let:

η:EF × F
xfx,gx.

By assumption, f and g are continuous, so by the universal property of the product topology, η is continuous. So eqf,g=η-1Δ is closed. ∎

Proposition 3.19.

Let E,𝒯E and F,𝒯F be two topological spaces. Let f:EF.

  1. If f open (i.e. U𝒯EfU𝒯F) and a surjection, and if the kernel of f is closed then F,𝒯F is Hausdorff.

  2. If f is a continuous, open surjection, then the kernel of f is closed if and only if F,𝒯F is Hausdorff.

Proof.

Let x,yF with xy. Since f is surjective, there exists w,zE such that fw=x and fz=y. Since xy, w,z is in the complement of the kernel of f, which is open by assumption. Since U × V:U,V𝒯E is a basis for the product topology on E2, there exists Uw,Uz𝒯E such that w,zUw × Uz and Uw × Uz is a subset of the complement of the kernel of f. Let tfUw and rfUz. If fr=ft then t,r lies in the kernel of f and in Uw × Uz, which is a contradiction. Hence Vx=fUw and Vy=fUz are disjoint. Since f is open, they are open. By construction, xVx and yVy. So F,𝒯F is Hausdorff.

Assume now that f is a continuous, open surjection. We just proved that since f is an open surjection, if its kernel is closed then F,𝒯F is closed. Conversely, if f is continuous and F,𝒯F is Hausdorff then its kernel is closed. Hence the equivalence stated. ∎

Remark 3.20.

We offer an alternative proof of the first assertion of the previous proposition when f is assumed to be open and bijective. We keep the notations used in that proposition.

Assume that f is an open bijection and kerf is closed. Since f is bijective, it has a right inverse g:FE. If U is open in E then g-1U=fU is open, so g is continuous. Moreover, letting Δ=y,y:yF and since f is surjective, we have:

Δ=fkernelf=g-1kernelf

which is therefore closed by assumption, as the preimage of the closed set kernelf by the continuous function g. Hence F,𝒯F is Hausdorff.

Remark 3.21.

An open map may not be continuous. For instance, a nonconstant map from 0,1 with the indiscrete topology to 0,1 with the discrete topology is always open but never continuous.

3.2. Limits along a set

Dealing with continuity can be complicated from the original definition. It is easier to introduce limits and the notion of continuity at a point. We first introduce a piece of vocabulary:

Definition 3.22.

Let E,𝒯E be a topological space. Let aE. The set of open neighborhoods of a in 𝒯E is the set V𝒯Ea of all U𝒯E such that aU.

Definition 3.23.

Let E,𝒯E and F,𝒯F be topological spaces. Assume F,𝒯F is a Hausdorff space. Let AE. Let aA¯ and lF. Let f:EF. We say that f has limit l at a along A when:

VV𝒯FlUV𝒯EafUAV.
Remark 3.24.

If aA¯ then the notion is silly, since we could find one open set U containing a and so that UA= in the above definition.

Proposition 3.25.

Let E,𝒯E and F,𝒯F be topological spaces. Assume F,𝒯F is a Hausdorff space. Let AE. Let aA¯ and lF. Let f:EF and g:EF. Assume that fx=gx for all xA. Then f has limit l at a along A if and only if g has limit l at a along A.

Proof.

Simply observe that the definition of limit involves only f(A) and g(A). ∎

We thus can define without ambiguity the limit of a partially defined function at a point in the closure of its domain:

Definition 3.26.

Let E,𝒯E and F,𝒯F be topological spaces. Assume AE is nonempty and f:AF. Then f has limit l at aA¯ along A if any extension of f to E has limit l at a along A.

Proposition 3.27.

Let E,𝒯E and F,𝒯F be topological spaces. Assume F,𝒯F is a Hausdorff space. Let BAE. Let aB¯ and lF. Let f:EF. Then if f has limit l at a along A then f has limit l at a along B.

Proof.

First, since BB, we have B¯A¯, so aB¯ implies that aA¯, hence the notion of limits are well-defined. Let VV𝒯Fl. Since limxa,xAfx=l, there exists UV𝒯Ea such that fUAV. Since BA we have fUBfUAV as desired. ∎

Proposition 3.28.

Let E,𝒯E and F,𝒯F be topological spaces. Assume F,𝒯F is a Hausdorff space. Let AE. Let aA¯ and lF. Let f:EF. If f has a limit at a along A then this limit is unique and denoted by limxa,xAfx.

Proof.

Assume f has limit l and l with ll at a along A. Since F is Hausdorff, there exists V,V open in F such that lV and lV. Then by definition of limits, there exists U,U open in E so that aU, aU and fUAV and fUAV. So fAUUVV=. This is absurd since aUU and aA¯ so AUU. ∎

Proposition 3.29.

Let E,𝒯E and F,𝒯F be topological spaces. Assume F,𝒯F is a Hausdorff space. Let AE. Let aA¯ and lF. Let f:EF. If f has a limit l at a along A then lfA¯.

Proof.

Let l=limxa,xAfx. Let V𝒯F such that lV. Then by definition of limit, there exists U𝒯E such that aU and fUAV. By definition, fUAfA. Since aA¯, we have AU so fAUVfA. So lfA¯. ∎

Remark 3.30.

In general, it is difficult to compute fA, so the result is used as follows: if fAB then limxa,xAfxB¯.

It is unpractical to check a statement for all open sets, as in general they are difficult to describe. The use of a basis makes things more amenable, and in fact it is the main role of basis.

Proposition 3.31.

Let E,𝒯E and F,𝒯F be topological spaces, with F,𝒯F Hausdorff. Assume that we are given a basis E for 𝒯E and a basis F for 𝒯F. Let AE, aA¯ and lF. Let f:EF. Let us denote by l the subset of all BF such that lB, and we define a similarly. Then f has limit l at a along A if and only if:

BlCafCAB.
Proof.

Assume 3.31. Let V𝒯F such that lV. By definition, there exists Bl such that lB and BV. By 3.31 there exists CmathcalBa, i.e. an open set in E containing a, such that fCABV. This shows that f converges to l at a along A.

Conversely, assume that f converges to l at a along A. Let Bl. By definition, B is an open set in F containing l so there exists U𝒯E with aU such that fUAB by definition of limit. There exists CE such that aC and CU as E is a basis for 𝒯E. Hence, fCAfCUC as desired. ∎

Proposition 3.32.

Let E,𝒯E and F,𝒯F be topological spaces, with F,𝒯F T1. Let f:EF, AE. If aA and if f has a limit at a along A then this limit is fa.

Proof.

Assume l is the limit of f at a along A and lfa. Then since 𝒯F is T1, there exists V𝒯F such that lV,faV. By definition of limit, there exists U𝒯E with aU and fUAV. Now, aAU so fafAUV which is absurd. ∎

Proposition 3.33.

Let E,𝒯E be a topological space. Let F,𝒯F be a Hausdorff topological space. Let A,AE. Let aA¯A¯. Then f has limit l at a along AA if and only if f has limit l at a along A and along A.

Proof.

Note that aA¯A¯ implies that aAA¯. The condition is necessary since AAA and AAA. Conversely, let VV𝒯Fl. By definition of limits along A and A there exist U,UV𝒯Ea such that fAUV and fAUV. Hence fUUAAV. Since UUV𝒯Ea, the proof is complete. ∎

Theorem 3.34.

Let E,𝒯E, F,𝒯F and G,𝒯G be three topological spaces, where 𝒯F and 𝒯G are Hausdorff. Let f:EF and g:FG. Let AE, aA¯. Set B=fA. If b is the limit of f at a along A and l is the limit of g at b along B then l is the limit of gf at a along A.

Proof.

Note first that bB¯, as necessary to make sense of the statement of the theorem. Let W𝒯G with lW. There exists V𝒯F with bV such that gVBW. Now, there exists U𝒯E such that aU and fUAV. By construction, fAUB so fAUBV. Hence, gfUAW as desired. ∎

Remark 3.35.

Beware of this theorem. Take fx=xsin1x, gx=0 for x0 and g0=1, and A=B=0 and a=b=0. Then:

0=limx0,x0fx=limy0,ynot-0gy

yet gf has no limit at 0 along the set of nonzero reals.

3.3. Continuity at a point

Definition 3.36.

Let E,𝒯E and F,𝒯F be topological spaces. Let aE. Then f is continuous at a if:

UV𝒯FfaUV𝒯EafUV.

The notion of continuity at a point looks more familiar when the codomain is Hausdorff.

Proposition 3.37.

Let E,𝒯E and F,𝒯F be topological spaces. Assume F,𝒯F is Hausdorff. Let aE. Then f is continuous at a if:

limxa,xEfx=fa

.

The main connection between continuity between spaces and continuity at a point is given in the following key theorem.

Theorem 3.38.

Let E,𝒯E and F,𝒯F be two topological spaces. The function f is continuous at every xE if and only if f is continuous on E.

Proof.

Assume first that f is continuous at every xE. Let V𝒯F. Let xf-1V. Then f is continuous at x so there exists Ux𝒯E such that xUx and fUxV (since fxV). Consequently, Uxf-1V. Hence:

f-1V=xf-1VUx

and thus f-1V is open. So f is continuous on E.

Conversely, assume f is continuous on E. Let xE. Let V𝒯F such that fxV. Then f-1V is open in E by continuity of f on E, and it contains x. Let U=f-1V. Then fUV. This concludes the proof. ∎

3.4. Limit of Sequences

We refer to the chapter Fundamental Examples for the topology on the one point compactification ¯ of .

Definition 3.39.

Let E be a set. A sequence in E is a function from to E.

Notation 3.40.

Sequences are denoted as families. Namely, if x is sequence in E, then we write xn for xn for n, and we usually identify x with xnn.

Remark 3.41.

By abuse of language, we will also call sequences functions from a subset n:nN of , for some N. The obvious re-indexing is left implicit.

The definition of limit for functions apply to sequences.

Definition 3.42.

Let E,𝒯 be a topological space. Let xnn be a sequence in E. Then we say that xnn converges to lE when the limit of xnn has limit l at along in 𝒯¯.

Remark 3.43.

We learnt that if two functions agree on some set A inside of a topological space E, then their limits along a agree (including whether they exist!). So the previous definition is understood as follows: choose any extension of a sequence to ¯ (i.e. pick some element x in E). Then take its limit at in ¯ along . Then the existence and, if applicable, the value of this limit is independant of the choice the extension of our sequence. It is the limit of the sequence.

Proposition 3.44.

Let E,𝒯 be a Hausdorff topological space. A sequence xnn converges to lE if and only if:

VV𝒯ElNnNxnV.
Proof.

Assume that xnn has limit l. Let VV𝒯El. By definition of limits and of the topology 𝒯¯, there exists a finite set S in such that xnV for all nF. Let N be the successor of the greatest element in F if F is nonempty, or 0 otherwise. Then for all nN we have xnV as desired.

The converse is obvious. ∎

Definition 3.45.

A subsequence of a sequence xnn in a set E is a sequence of the form xϕnn for some strictly increasing function ϕ:.

Lemma 3.46.

Let ϕ: be strictly increasing. Then for all n we have ϕnn.

Proof.

By definition, ϕ00. Assume that for some n we have ϕnn. Then by assumption, ϕn+1>ϕnn so ϕn+1n+1. The lemma holds by the theorem of induction. ∎

Proposition 3.47.

Let E,𝒯 be a topological space. If a sequence xnn in E converges to l then all subsequences of xnn converge to l.

Proof.

Assume that xnn converges to l and let ϕ: be a strictly increasing function. Let VV𝒯l. By assumption, there exists N such that for all nN we have xnV. Therefore, xϕnV since ϕnn. Hence the subsequence xϕnn converges to l as well. ∎

Theorem 3.48.

Let E,𝒯 be a topological space. A sequence xnn in E converges to l if and only if every subsequence of xnn has a subsequence which converges to l.

Proof.

The condition is necessary by the previous result. Let us show it is sufficient. Assume that xnn does not converge to l. Then there exists VV𝒯l such that for all N there exists nN such that xnV. Set ϕ0 to be the smallest n such that xnV. Assume we have constructed ϕ0<<ϕN for some N such that xϕkV for k=0,,N. Then let AN=n:nϕN+1. Then AN, so it has a smallest element ϕN+1. Note that by construction, ϕN+1>ϕN and xϕN+1V. Hence we have constructed a subsequence xϕnn entirely contained in EV. By construction, it has no subsequence which converges to l. ∎

4. Limits and Continuity in Metric Spaces

In this section, the topology on any metric space is meant as the metric topology.

4.1. Convergence in metric spaces

Proposition 4.1.

Let E,d be a metric space. A sequence xnn in E converges to l if and only if:

ε>0NnNdxn,l<ε.
Proof.

Since open balls form a basis for the metric topology, the definition of limit for a sequence at infinity (along ) is equivalent to:

ε>0NnNxnBl,ε

which is equivalent by definition to the statement in the proposition. ∎

4.2. Squeeze Theorems

The following result is often useful.

Theorem 4.2.

Let E,d be a metric space. Let xnn be a sequence in E and lE. If there exists a sequence rnn in 0, and N such that limnrn=0 and dxn,lrn for all nN, then limnxn=l.

Proof.

Let varepsilon>0 be given. Since rnn converges to 0 in 0,, there exists N such that for all nN, rn0,ε. Hence, for nN we have dxn,lε, as desired. ∎

This theorem shows that a decent supply of sequences converging to 0 in is useful. The following results help in these constructions.

Proposition 4.3.

Let φ:0, be increasing and not bounded. Let M0,. Then Mφnn converges to 0.

Proof.

Let ε>0 be given. By assumption, there exists N such that φn>Mε. Since φ is increasing, for all nN we have φnMε. Hence for all nN we have MφnεMM=ε, as desired. ∎

The existence of unbounded functions φ is a consequence of the fact that is Archimedean. Namely, the injection i: is unbounded precisely because of the Archimedean principle (in fact, the two statements are equivalent). Therefore, 1nn* converges to 0, as desired.

Theorem 4.4.

Let xnn, ynn and znn be three sequences in a linearly ordered set (E,), and N such that for all nN we have xnynzn. If xnn and znn have limit l then so does ynn.

Proof.

The above result is used in under many names: squeeze theorem, guardsmen theorem, pinching theorem, and more. It is often useful to prove limits.

Of course, is a field, and we will have to address the matter of the continuity of the operations. We shall do this later, however, as we do not need them at this moment.

4.3. Closures and Limits

Theorem 4.5.

Let E,d be a metric space. Let AE. Then aA¯ if and only if there exists a sequence xnn in A converging to a.

Proof.

If there exists a sequence in A converging to a then aA¯.

Conversely, assume aA¯. For n*, there exists xnB(a.1n)A. By construction, dxn,a1n for all n* so xnn* converges to a, which completes our proof. ∎

Limits of subsequences can be characterized in metric spaces as follows:

Theorem 4.6.

Let E,d be a metric space. Let xnn be a sequence in E. Then lE is a limit for a subsequence of xnn if and only if:

lnxk:kn¯.
Proof.

Assume l is the limit of some subsequence xϕnn of xnn. Let n. Then l is the limit of the truncated sequence k:knxϕk at , so lxk:kn¯ since ϕnn. Hence the condition is necessary.

Assume now lnxk:kn¯. To ease notations, write Xn=xk:kn for all n. Since lX0¯, the set k:dxk,l<1 is not empty. Let ϕ0 be its smallest element.

Assume now we have constructed ϕ0<<ϕn for some n such that dxϕk,l<1k+1 for k=0,,n. Since lXϕn+1¯, the set k:k>ϕndxϕk,l<1n+2 is not empty. Let ϕn+1 be its smallest element.

One checks that l=limnxϕn as desired. ∎

4.4. Limits at a point

Theorem 4.7.

Let E,dE, F,dF be metric spaces. Let f:EF be a map. Let xE, AE with xA¯, and lF. Then f is has limit l at x along A if and only if for any sequence xnnN in A which converges to x, we have limnfxn=l.

Proof.

Note that since xA¯ and E,d is metric, there is a sequence of elements of A converging to x. The theorem on composition of limits shows that the condition is necessary. Let us show that it is sufficient by contraposition. Assume f does not converge to l at x. Hence, there exists ϵ>0 such that for all δ>0, there exists zE such that dEx,z<δ yet dFfz,l>ϵ. For each n*, pick xnA such that dExn,z<1n and dFfxn,l>ϵ. Note that since xA¯, we can always choose such an xn. By the squeeze theorem, we conclude that limnxn=x. On the other hand, fxnn* does not converge to l. This proves our result. ∎

4.5. Continuity at a Point

Theorem 4.8.

Let E,dE, F,dF be metric spaces. Let f:EF be a map. Let xE. Then f is continuous at x if and only if for any sequence xnnN which converges to x in E, we have limnfxn=fa.

Proof.

This is a direct application of the result on limits in a metric space. ∎

5. Filters

This chapter introduces filters as a mean to analyze topological spaces. In a sense to be illustrated below, filters provide a generalization of sequences, albeit in a dual manner.

5.1. Filters on sets

Definition 5.1.

Let E be a set. A filter base on E is a subset of 2E such that:

  1. is not empty and,

  2. If A,B then there exists C such that CAB.

Definition 5.2.

Let E be a set. A filter on E is a subset of 2E such that:

  1. ,

  2. E,

  3. If A and CE with AC then C,

  4. If A,B then AB.

Hence a filter is a proper nonempty upset closed under finite intersections. Note that the empty set is never in a filter. Note that a filter is a filter basis.

Proposition 5.3.

Let be a filter basis on a set E. Let:

=AE:BBA.

Then is a filter on E, and it is the smallest filter on E containing . We call it the filter generated by .

Proof.

By definition, so is not empty. Moreover, if A and CE such that AC then, by construction, there exists B such that BA and thus BC so C. If A1,A2 then there exists B1,B2 such that BiAi for i=1,2. Thus B1B2A1A2. Since is a filter basis, there exists C such that CB1B2 so CA1A2. Hence A1A2. Last, if then there must be A such that A, i.e. which is a contradiction. So . So is a filter.

Let G be some filter containing . Let A. Then by definition, there exists B such that BA. Sicne BG and G is a filter, we conclude AG. Hence G as desired. ∎

Proposition 5.4.

Let E, F be sets and f:EF be a function. If is a filter basis on E, then f, defined as:

f=fA:A

is a filter basis.

Proof.

By definition, f is not empty and does not contain the emptyset. Now, let A,Bf. By definition, there exists C,D such that A=fC and B=fD. Now, since is a filter basis, there exists H such that HCD. Hence fHAB, and by definition, fHf. ∎

Example 5.5.

Let f:EF be constant, equal to tF, while E,F have more than one element. Let xE. Then =AE:xA is a filter on E. Now, f=t. One checks easily that this is a filter basis, but not a filter (as it would contain F which is assumed to contain a set of cardinal 2). Hence, images of filters may not be filters. We could define f to be the filter generated by the filter basis f, thus defining a map from filters to filters for each map from E to F.

5.2. Limits of Filters basis

Definition 5.6.

Let E,𝒯 be a topological space. Let xE. Let be a filter basis on E. We say that converges to x when:

U𝒯(xU)BBU.
Remark 5.7.

Equivalently, a filter basis converges to x if the set V𝒯x of open neighborhoods of x is contained in the filter generated by .

Example 5.8.

The set V𝒯x is a filter basis which converges to x. The elements of the filter generated by V𝒯x are called neighborhoods of x.

The following theorem illustrates the connection between filters and sequences.

Theorem 5.9.

Let E,𝒯 be a topological space. Let xnn be a sequence in E. For n we define Xn=xk:kn. Then X=Xn:n is a filter basis. Moreover, xnn converges to x if and only if X converges to x.

Proof.

Since Xn+1Xn for all n, it is immediate that X is a filter basis. Assume that xnn converges to x. Let U𝒯 such that xU. Then by definition, there exists N such that for all nN we have xnU. Hence XNU. Thus, X converges to x (since U was arbitrary). Conversely, assume that X converges to x. Let U𝒯 such that xU. By definition, there exists XNX such that XNU. Hence for all nN we have xnU. So xnn converges to x (since U was arbitrary). ∎

Remark 5.10.

Let xnn be a sequence. Let xφnn be a subsequence. The filter generated by the filter basis xk:kn:n is a subset of the filter generated by the filter basis xϕk:kn:n. So filters “grow” as we take subsequences. From this observation, we note that by definition, it is now immediate to see that if a sequence converge to some x then all of its subsequences do too. The following proposition generalizes this observation.

Proposition 5.11.

Let E,𝒯 be a topological space. Let and G be two filter bases on E such that for all F there exists GG such that GF (we say that G is finer than ). If converges to x, then G converges to x.

Proof.

Let x be a limit of . Let U𝒯 such that x𝒯. There exists F such that FU. By assumption, there exists GG such that GF. Hence G converges to x. ∎

Corollary 5.12.

Let E,𝒯 be a topological space. If G are two filters on E and if converges to x, then G converges to x.

Limits of filters may not be unique. We have:

Theorem 5.13.

Let E,𝒯 be a topological space. The following are equivalent:

  1. The space E,𝒯 is Hausdorff,

  2. Every convergent filter basis on E has a unique limit.

Proof.

Assume that E,𝒯 is Hausdorff. Let be a filter basis on E. Assume it converges to x and y in E. If xy then, since 𝒯 is Hausdorff, there exists Ux,Uy𝒯 such that xUx, yUy and UxUy=. By definition of convergence, there exist Bx,By such that BxUx and ByUy. By definition of filter basis, there exists C such that CBxBy=. This contradicts the fact , as a filter basis, does not contain the empty set. Hence limits are unique if they exist.

Assume now that all filter bases have unique limits in E,𝒯. In particular, for any xE, the filter basis V𝒯x of open neighborhoods of x in 𝒯, has x for unique limit. Let yx. Suppose that for all UV𝒯y, for all VV𝒯x, we have UV. Let:

=UV:UV𝒯x,VV𝒯y.

By assumption, is a nonempty set of elements in 𝒯 not containing the emptyset. Moreover, it is closed by finite intersection (as 𝒯 is). Hence, it is a filter basis. Yet, for any UV𝒯x and any VV𝒯y, we have U,V so converges to x and y by definition. This is a contradiction. Hence, for all UV𝒯y there exists VV𝒯x such that UV=, i.e. E,𝒯 is Hausdorff. ∎

Notation 5.14.

When E,𝒯 is Hausdorff and is a filter basis converging to x then we write: x=lim.

5.3. Some Applications of Filters

As a rule, filters can be used to replace sequences in general topology to obtain results limited to metric spaces. For instance:

Lemma 5.15.

Let E,𝒯 be a topological space. Let AE. Let be a filter basis in A which converges to x in 𝒯. Then xA¯.

Proof.

Let VV𝒯x. By definition of convergence, there exists B such that BV. Now, BA by assumption, so emptysetBVA. So xA¯ since V is arbitrary. ∎

Theorem 5.16.

Let E,𝒯 be a topological space. Let AE. The closure of A in 𝒯 is the set of all limits in E of filters of A.

Proof.

All limits of filters of A are in A¯ by the previous lemma. Assume xA¯. If UV𝒯x then AU. It is then obvious that:

UA:UV𝒯x

is a filter basis of A which converges to x. ∎

Definition 5.17.

Let E be a set, F,𝒯 be a topological space, a filter basis on E, and f:EF. Then we say that f converges to xF along if f converges to x.

When F,𝒯F is Hausdorff, we shall write: limf for this limit.

Theorem 5.18.

Let E,𝒯E and F,𝒯F be two topological spaces, and let f:EF. Let AE, aA¯. Then f has limit l at a along A if and only if for all filter bases of A converging to a, the filter basis f converges to l in F.

Proof.

Assume first that f converges. Let be a filter basis of A converging to a. Note that we have shown such a filter basis exists. Now, let U𝒯F such that lU. Since f converges to l at a along A, there exists V𝒯E with aV such that fAVU. Since converges to a, there exists C such that CV. Since CA by assumption, C=CAVA so fCU as desired. Since U was arbitrary, f converges to l as desired.

Conversely, assume that f converges to l for all filter bases of A converging to a. Let U𝒯V such that lU. The filter basis V=VA:V𝒯,aV converges to a and is a subset of 2A so by assumption, fV converges to l. Thus, there exists CV such that fCU. Hence, by definition, there exists V𝒯E such that fVAU. As U is arbitrary in V𝒯Fl, we conclude that f converges to l at a along A. ∎

Corollary 5.19.

Let E,𝒯E and F,𝒯F be topological spaces, xE and f:EF. Then f is continuous at x if and only if the image by f of all filter bases of E converging to x converge to fx.

5.4. Ultrafilters

Definition 5.20.

Let E be a set. An ultrafilter on E is a filter such that for all AE, we have either A or EA.

Example 5.21.

Let xE. Then AE:xA is an ultrafilter.

Since for no set A can we have A and EA in a filter of E, this definition suggests the following:

Theorem 5.22.

A filter on a set E is an ultrafilter if and only if it is maximal among all filters on E (for the inclusion).

Proof.

Assume first that is an ultrafilter on E. Let G be a filter on E such that G. Let AG. Assume EA. Then EAG, and thus =AEA is in G, as filters are closed under finite intersections. This is a contradiction. Since is an ultrafilter, A. As A was arbitrary, G=.

Conversely, assume that is a maximal filter for inclusion. Assume that it is not an ultrafilter. Let YE such that Y,EY. Without loss of generality, assume Y. Let:

G=CE:AAYC.

Now, by construction, G is not empty. Assume G. Then there exists A such that AY=. Thus AEY. As is a filter, this would imply EY which is a contradiction. So G does not contain the empty set. By construction, if AG and AC then CG. Moreover, if A1,A2G then there exist B1,B2 such that BiYAi for i=1,2. Thus B1B2YA1A2. Since B1B2 we conclude that A1A2G. Thus G is a filter. By definition, since AYA for all A, we have G. Since EY=Y and E, we have YG, yet Y. This contradicts the maximality of . So is an ultrafilter. ∎

Maximality lends itself to the application of Zorn’s lemma to prove the existence of ultrafilters containing any given filer.

Theorem 5.23.

Let E be a set, and let be a filter basis on E. Then there exists an ultrafilter G containing .

Proof.

Since every filter basis is contained in a filter, we will assume is a filter. Let:

𝒫=Hfilters on E:H.

The set 𝒫 is ordered by inclusion, and it is not empty (it contains . Note that 𝒫 is an upper set, and thus a maximal element of 𝒫 is a maximal element of the set of all filters on E. Thus, it is enough to show that 𝒫 has a maximal element.

Let 𝒰 be a chain in 𝒫, i.e. a linearly ordered subset of 𝒫. Let Y=𝒰. If A,BY then there exists U,V𝒰 such that AU and BV. Since 𝒰 is linearly ordered, we may assume UV. Thus A,BV, and V is a filter, so ABVY. The set Y is not empty (filters never are), does not contain the empty set (filters never do), and is an upper set: if AY and AC for some C, then there exists U𝒰 such that AU, and since U is a filter, CU so CY. Therefore, Y is a filter. Trivially Y𝒫. Hence, 𝒰 admits an upper bound. Since 𝒰 is an arbitrary chain, Zorn’s lemma implies that 𝒫 admits a maximal element G. Thus G is a maximal filter, and hence an ultrafilter, containing . ∎

Due to this result, it is possible to rephrase the theorems in the application section of this chapter in terms of ultrafilters only.

6. Compactness

6.1. Compact space

Definition 6.1.

A topoligical space E,𝒯 is compact if, given any 𝒰𝒯 such that E=𝒰, there exists a finite subset V of 𝒰 such that E=V.

Remark 6.2.

A subset 𝒰 of 𝒯 such that E=𝒰 is called an open covering of E.

Theorem 6.3.

Let E,𝒯 be a topological space. Then the following are equivalent:

  1. E,𝒯 is compact,

  2. For any set of closed subsets of E whose intersection is empty, there exists a finite subset G of whose intersection is empty.

Proof.

The theorem follows by taking complements. ∎

Corollary 6.4.

Let E,𝒯 be a compact space. Let Fnn be a decreasing sequence of nonempty closed subsets of E (where the order on sets in inclusion). Then:

nFn.
Proof.

If nFn= then since E,𝒯 is compact, we can find N such that F1F2FN=. Yet since the family is decreasing, F1F2FN=FN which is assumed not empty. Our result follows by contraposition. ∎

Theorem 6.5.

Let E,𝒯 be a topological space. Let be a set of closed subsets of E such that given any finite G, we have G (we say that has the finite intersection property). Then if and only if E,𝒯 is compact.

Proof.

Assume first that E,𝒯 is compact and consider a set of closed subsets of E. Assume =. Since E is compact, there exists a finite set G such that G=. This proves the necessity of our theorem by contraposition.

Conversely, assume that for all sets of closed subsets of E with the finite intersection property, . Let be a set of closed subsets of E such that =. Then by our assumption, there exists at least one finite subset G such that G=. ∎

Example 6.6.

The space E,2E is compact if and only if E is finite.

Example 6.7.

It is striaghtforward that if Un=-n, in for all n then Un:n is an open covering of which has no finite subcovering. So is not compact.

6.2. Compact Subspaces

Definition 6.8.

Let E,𝒯 be a topological space. Let AE. Then A is a compact subspace of E,𝒯 if A,𝒯A is a compact space (where 𝒯A is the trace topology on A).

Remark 6.9.

We will say that A is compact in E when A is a compact subspace of E.

Theorem 6.10.

Let E,𝒯 be a topological space. Let AE. Then A is compact if and only if for all 𝒰𝒯 such that A𝒰, there exists n and U1,,Un𝒰 such that:

Ak=1nUk.
Proof.

Assume that A,𝒯A is compact in E,𝒯. Let 𝒰𝒯 such that A𝒰. Then A=UA:U𝒰. By definition, UA𝒯A for all U𝒯. Hence by compactness of A, there exists U1,,Un𝒰 such that A=i=1nAUi. Hence Ai=1nUi.

Assume now that any open covering of A in E admits a finite covering. Let V𝒯A such that A=V. By definition of 𝒯A, for each VV there exists UV𝒯 such that V=UVA. Thus AUV:VV, so there exists UV1,,UVn𝒯 such that Ai=1nUVi. Hence A=i=1nVi. ∎

Theorem 6.11.

Let E,𝒯 be a topological space. Let AE. Then A is compact if and only if for any family FiiI of closed sets in E such that:

AiIFi=

there exists a finite subset JI such that:

AiJFj=.
Proof.

Take complements in previous theorem. ∎

Theorem 6.12.

Let E,𝒯 be a Hausdorff topological space. Then if AE is compact, then A is closed.

Proof.

Let xEA and yA. Since 𝒯 is Hausdorff, there exists Ux,y,Vx,y𝒯 such that Ux,yVx,y=, with yUx,y and xVx,y. Let 𝒰x=Ux,y:yA. By construction:

A𝒰x

and thus, since A is compact, there exists n and y1,ldots,ynA such that:

Ak=1nUx,yk.

Let:

Wx=k=1nVx,yk.

As a finite intersection of open sets, Wx is open. Moreover:

Ak=1nUx,yk
k=1nEVx,yk
=Ek=1nVx,yk=EWx.

Hence, WxEA. Therefore:

EA=xAWx

and thus EA is open, so A is closed as desired. ∎

Theorem 6.13.

Let E,𝒯 be a compact space. Then if AE is closed, then it is compact.

Proof.

Let 𝒰𝒯 such that A𝒰 and let A be closed. Then EAtau so, if V=𝒰EA, then E=V. Since E,𝒯 is compact, there exists a finite set WV such that E=W. Now, if Z=WEA, one checks readily that AZ and by construction, Z𝒰 with Z finite. ∎

Corollary 6.14.

Let E,𝒯 be a compact Hausdorff space. Then AE is compact if and only if A is closed.

6.3. Continuous image of a compact space

Theorem 6.15.

Let E,𝒯E be a compact space, F,𝒯F be a topological space, and f:EF be a continuous function. Then fE is a compact subspace of F.

Proof.

Let 𝒰𝒯F such that fE𝒰. Since f is continuous, for all U𝒰, we have f-1U𝒯E. Let V=f-1U:U𝒰. Then V𝒯E and by construction, V=E. Since E,𝒯 is compact, there exists nN and U1,,Un𝒰 such that E=k=1nf-1Uk. Hence fEk=1nUk as desired. ∎

Corollary 6.16.

Let E,𝒯E and F,𝒯F be topological spaces. If f:EF is continuous, and AE is compact, then fA is compact.

Theorem 6.17.

Let E,𝒯E be a compact space, and F,𝒯F be a Hausdorff topological space. Let f:EF be a continuous bijection. Then f is a homeomorphism.

Proof.

It suffices to show that f-1 is continuous. Let AE be a closed set. Then A is compact since E is compact. So fA is compact in F. Since F is Hausdorff, fA is closed. Since f is a bijection, f-1-1A=fA. So f-1 is continuous. ∎

6.4. Compact Metric Spaces

Theorem 6.18.

Let E,d be a metric space. Then the following are equivalent:

  1. E is compact in its metric topology.

  2. Bolzanno-Weierstrass Every sequence in E admits a convergent subsequence.

Proof.

First, assume that E is compact. Let xnn be a sequence in E. Let Xn=xk:k,kn¯ for all n. The sequence Xnn is a decreasing sequence of nonempty closed subsets of E, which is compact, so nXn. By Theorem (LABEL:AccumulationSeqThm), this implies that xnn admits a convergent subsequence.

Second, assume that the Bolzanno-Weierstass axiom holds for E. We first prove a few lemmas.

Lemma 6.19 (Lebesgues number).

Let 𝒰𝒯d be an open covering of E. There exists ϵ>0 such that for all xE, there exists U𝒰 such that the open ball Bx,ϵ of center x and radius ϵ is contained in U.

Proof.

Assume that for all ϵ>0, there exists xE such that for all U𝒰, we have Bx,ϵU. Let n. There exists xnE such that Bxn,1n+1U for all U𝒰. The sequence xnn admits a convergent subsequence xϕnn by the Bolzanno-Weierstrass axiom. Let lE be its limit. Since 𝒰 covers E, there exists U𝒰 such that lU. Since U is open, there exists δ>0 such that Bl,δU. Since xϕnn converges to l, there exists N2 such that for all nN2, we have xϕnBl,δ2. Let N such that 1N+1<δ2 and NN2. Let zBxϕN,1ϕN+1. Then:

dz,ldz,xϕN+dxϕN,l<1ϕN+1+δ2<δ

since ϕNN. Hence, zBl,δ so BxϕN,1ϕN+1Bl,δU. This is a contradiction. ∎

Lemma 6.20 (Precompact).

For any ϵ>0 there exists n and x1,,xnE such that:

E=j=1nBxj,ϵ.

.

Proof.

Assume that there exists ϵ>0 such E is not the union of finitely many open balls of radius ϵ. Let x0E: since EBx0,ϵ, there exists x1E such that dx0,x1ϵ.

Assume we have constructed x0,,xnE such that dxi,xjϵ for ij0,,n. Since Ej=0nBxj,ϵ, there exists xn+1Ej=0nBxj,ϵ, i.e. dxn+1,xiϵ.

By induction, we have constructued a sequence xnn which admits no convergent subseqence. This contradicts the Bolzanno-Weierstrass axiom. ∎

We can now conclude our proof. Assume 𝒰 is an open covering of E. There exists ϵ>0 such that any open ball of radius ϵ or less is contained in some U𝒰. Now, there exists x1,,xnE such that E=j=1nBxj,ϵ. Let Uj𝒰 such that Bxj,ϵUj. Then E=j=1nUj. So E is compact, as desired. ∎

Corollary 6.21.

A compact metric space is bounbed.

Proof.

The precompact lemma shows that a compact metric space can be covered with finitely many balls of radius 1. ∎

Remark 6.22.

The definition of compacity by means of open covering is referred to the Heine-Borel axioms of compactity. Thus we have shown that in metric spaces, Heine-Borel and Bolzanno-Weierstrass are equivalent.

6.5. Compacts of

We provide two proofs of the following result, both based on your real analysis class.

Theorem 6.23.

A subset A of is compact if and only if it is closed and bounded.

Necessity.

Since is Hausdorff, compact subsets are closed. Since is metric, compact subsets are bounded. ∎

First proof of sufficiency: monotonicity.

Assume that A is closed and bounded. Let xnn be a sequence in A. By the monotone subsequence theorem, there exists a monotone subsequence of xnn. Since A is bounded, this subsequence must converge. Hence A satisfies Bolzanno-Weierstrass axiom in a metric space and is thus compact. ∎

Second proof of sufficiency: dichotomy.

Let a0<b0 be real numbers. Let xnn be a sequence in a0,b0. Set ϕ0=0.

Assume that for some n we have constructed ϕ0<<ϕn, a0a1anb0 and a0bnb0 such that xϕkak,bk and bk-ak=12kb0-a0. Let mn=an+bn2. If n:nϕnxnan,mn is infinite, then let ϕn+1 be its smallest element. We also let an+1=an, bn+1=mn. Otherwise, we set ϕn+1 to be the smallest element of the (necessarily infinite) set n:nϕnxnmn,bn, and we let an+1=mn, bn+1=bn. One checks that by induction, we have constructed an subsequence xϕnn and monotone sequences ann and bnn such that:

  1. ann is increasing and bounded above by b0 so it converges.

  2. bnn is decreasing and bounded above by a0 so it converges.

  3. We have limnan-bn=0 so limnan=limnbn.

  4. Thus by the squeeze theorem, limnxϕn=limnan.

Hence again we have proven the Bolzanno-Weierstrass axiom for all closed intervals. Now, if A is closed and bounded, then it is a closed subset of some closed interval, i.e. of a compact space. Hence it is compact, as needed. ∎

Remark 6.24.

The second proof is longer but can be applied as is to prove that compact subsets of n are closed bounded subsets of n. Indeed, rather than cut an interval in two, one can repeat the argument, cutting a square in four, a cube in eight, and so forth. This process works well as long as we divide an hypercube into finitely many hypercubes at each stage. This proofs fails for infinite dimensional normed vector spaces, and in fact so does the result: one can show that locally compact normed vector spaces are all finite dimensional.

6.6. Uniform Continuity and Heine Theorem

Definition 6.25.

Let E,dE and F,dF be metric spaces. A function f:EF is uniformly continuous when:

ϵ>0δ>0x,yEdEx,y<δdFfx,fy<ϵ.
Proposition 6.26.

A uniformly continuous function is continuous.

Proof.

Obvious. ∎

Example 6.27.

A L-Lipshitz function is uniformly continuous: take δ=ϵL if L>0 (i.e. nonconstant functions), and δ arbitrary for constants.

Example 6.28.

The function xx2 is not uniformly continuous in . Indeed basic precalculus shows that x2<kx if and only if x-k,k.

The following shows that continuity can be strengthened to uniform continuity when working on a compact metric space.

Theorem 6.29.

Let E,dE be a compact metric space. Let F,dF be a metric space. Let f:EF be given. Then f is continuous if and only if f is uniformly continuous on E.

Proof.

If f is uniformly continuous then it is continuous. Let us prove the converse. Assume f is continuous on E. Let ϵ>0. Then for all xE there exists δx>0 such that if yBx,δx then dFfy,fx<ϵ2. Now, since E=xEBx,δx2, and E is compact, there exists x1,,xn such that E=j=1nBxj,δxj2. Let δ=12minδxj:j=1,,n>0. Let x,yE with dEx,y<δ. There exists j1,,n such that xBxj,δxj. Then

dEy,xjdEy,x+dEx,xj<δ+12δxjδxj.

Thus:

dFfx,fydFfx,fxj+dfxj,fy<ϵ

as desired.

Alternate, use real analysis..

If f is uniformly continuous then it is continuous. Let us prove the converse. Assume f is continuous but not uniformly continuous. There exists ϵ>0 such that for all δ>0 there exists x,yE such that dEx,y<δ and dFfx,fyϵ. For n let xn,ynE such that dExn,yn<1n+1 and dFfxn,fynϵ. Since E is compact, there exists a convergent subsequence xϕnn. Then there exists a convergent subsequence yϕψnn of yϕnn. Let lx=limnxϕψn and ly=limnyϕψn. Now lx-ly=limnyϕ(ψ(n)))-xϕψn=0. On the other hand, using the continuity of f, we have ϵdFflx,fly which is a contradiction. ∎

Remark 6.30.

The notion of completeness of metric spaces is not topological: there exist two homeomorphic metric spaces, one complete and one not. A possible choice of morphisms for the category of metric spaces are the uniformly continuous functions, in which case isomorphic metric spaces are either all complete or all not complete.

Now, if E and F are compact metric spaces, and if they are homeomorphic, then E is complete if and only if F is complete. This is a very remarkable fact: a purely topological notion (compactness) imposes a metric notion (completeness). This is seen by observing that a Cauchy sequence has at most one accumulation point, and we have seen that every sequence of a compact space has at least one. With the Heine theorem, we see a different reason: an homeomorphism between compact metric spaces is always bi-uniformly continuous, and hence preserve Cauchy sequences.

6.7. Filter characterization of compactness

Theorem 6.31.

Let E,𝒯 be a topological space. E,𝒯 is compact if and only if every ultrafilter in E converge.

Proof.

Assume that E,𝒯 is compact. Let be an ultrafilter in E. Assume it does not converge. Then for all xE there exists Ux𝒯 with xUx such that Ux. As is an ultrafilter, EUx. Now, E=xEUx, and as 𝒯 is compact, there exists a finite subset F of E such that E=xFUx. Now, =xFEUx which is a contradiction. So converges.

Assume now that all ultrafilters in E converge in 𝒯. Assume 𝒯 is not compact. Let W𝒯 such that E=W, with W infinite, and such that no finite subset of W covers E. Let:

=EUFU:FW,finite.

Since no finite subset of W covers E, the emptyset is not in . By construction, is closed by finite intersections, and thus it is a filter basis. Assume it converges to x. Since W covers E, there exists WW such that xW. By definition of convergence, there exists a finite set FW such that EUFUW, i.e.:

E=UFWU

with FW finite subset of W. This is a contradiction. Hence does not converge. Now, is contained in some ultrafilter G, which therefore does not converge at all. ∎

6.8. Tychonoff theorem

Theorem 6.32.

Let Ei,𝒯iiI be a family of compact topological spaces. Let 𝒯 be the product topology of 𝒯iiI. Then iIEi,𝒯 is compact.

Proof.

Let be an ultrafilter in iIEi. Let iI. Let pi:jIEjEi be the canonical surjection. Let i=pi. Now, let Ai and CEi such that CAi. Since is a filter, and since Api-1Api-1C, we conclude pi-1A and pi-1C. Since pi is a surjection, C=pipi-1Ci by definition. It is also immediate that i is a nonempty subset of 2Ei to which the empty set does not belong. Now, the same reasonning shows that if A,Bi then ABi. Hence, i is a filter in Ei. Last, let CEi. Then either pi-1C or its complement are in , since is an ultrafilter. Since C=pipi-1C as pi is a surjection, either C or its complemenent belongs to i. Thus i is an ultrafilter on Ei.

Since Ei is compact, i converges to some xi.

Let us show that converges to xiiI in the product topology 𝒯. Let U𝒯 such that xiiIU. Let W𝒯 and JI finite such that piW=Ei for iJ, and WU, xiiIW. Write Wi=piW for all iI and note that xiWi, with Wi𝒯i for all iI. Also note that W=jFpj-1Wj. By definition of convergence and xiiI, we have Wii, so pj-1Wj. Since is closed under finite intersections, we conclude W, hence U. So converges to xiiI as desired. ∎

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