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\begin{document}

\noindent
{\large \bf Additional notes \#2.}

\bigskip

\noindent
{\bf Paths.}

\bigskip

Let $\lb X; T\rb$ be a topological space.
For $a, b\in X$ write $a\stackrel{f}{\leadsto} b$
to mean that $f$ is a path from $a$ to $b$.
(That is, $f: [0,1]\rightarrow X$ is a continuous function
satisfying $f(0) = a$ and $f(1) = b$.)

\bigskip

\noindent
{\bf Theorem}. If $a\stackrel{f_1}{\leadsto} b$
and $b\stackrel{f_2}{\leadsto} c$, then 
$a\stackrel{f}{\leadsto} c$ where $f$ is defined by
$$
f(r) = \left\{\begin{array}{ll}
              f_1(2r)&\mbox{\rm if $0\leq r\leq 1/2$}\\
              f_2(2r-1)&\mbox{\rm if $1/2 \leq r\leq 1$.}
\end{array}\right.
$$
\medskip

{\it Proof.} We must show that $f$ is a continuous
function from $[0,1]$ to $X$ satisfying 
$f(0) = a$ and $f(1) = c$. Each part of this claim
is clear, except that $f$ is uniqely defined
at $r = 1/2$ and that $f$ is continuous.
The fact that $f$ is uniquely defined at $r = 1/2$
follows from the fact that $f(1/2) = f_1(2\cdot(1/2)) = 
f_1(1) = b$ in the first case of the definition of $f$,
and $f(1/2) = f_2(2*(1/2)-1) = f_2(0) = b$
in the second case of the definition. Thus $f(1/2)$
is uniquely defined to be $b$.
To show that $f$ is continuous, we apply our continuity test:

\medskip

\noindent
{\sl Continuity Test:} A function $g: Y\rightarrow Z$
is continuous iff it is continuous at each $y\in Y$.
This means that for each $y\in Y$ and 
each basic open set $U\in T(Z)$ for which 
$f(y)\in U$ there is a basic open set $V\in T(Y)$
such that $y\in V$ and $g(V)\subseteq U$.

\medskip

For this proof we will always use the following type
of basis for a subspace $[a,b]\subseteq {\BR}$
that is a closed interval: 
$${\mathcal B} = \{ [a,t)\;|\;a < t < b\}\cup
\{ (s,t)\;|\;a < s < t < b\}\cup
\{ (s,b]\;|\;a < s < b\}.$$

Let $\widehat{f}_1(r) = f_1(2r)$ and let $\widehat{f}_2(r) =
f_2(2r-1)$.
Note that $\widehat{f}_1: [0,1/2]\rightarrow X$
is the composition of two continuous functions,
$f_1$ and the doubling function from
$[0,1/2]$ to $[0,1]$, so $\widehat{f}_1$
is continuous. A similar argument shows that 
$\widehat{f}_2$ is continuous. We prove
that $f$ is continuous in two cases.

\medskip

\noindent
{\bf Case 1.} $f$ is continuous at any $r\neq 1/2$.

Assume that $r\in [0,1/2)$. Then 
for some $\epsilon >0$
the set $O = (r-\epsilon, r+\epsilon)$
is contained in $[0,1/2)$. 
On $O$ the function 
$f$ agrees with $\widehat{f}_1$. 
Now suppose that $U$ is a basic open set in $T(X)$
and $f(r) \in U$. Then $\widehat{f}_1(r) = f(r)\in U$,
so there is a basic open set $V\in T([0,1/2])$
such that $\widehat{f}_1(V)\subseteq U$.
By shrinking this basic open set $V$ if necessary 
we may assume it has the form $(r-\delta,r+\delta)$
for some $\delta$ satisfying $0 < \delta < \epsilon$.
Thus $V\subseteq O$, and this means that $f$ agrees
with $\widehat{f}_1$ on $V$. Hence
$f(V)\subseteq U$. Since $V$ is also a basic open
in $T([0,1])$, it follows that $f$
is continuous at $r$.

If $r\in (1/2,1]$, then the proof is the same as
the above using $\widehat{f}_2$ in place of $\widehat{f}_1$.

\medskip

\noindent
{\bf Case 2.} $f$ is continuous at $r = 1/2$.

Let $U$ be a basic open set in $T(X)$ containing
$f(1/2) = b$. Since $\widehat{f}_1$ is continuous,
and $\widehat{f}_1(1/2) = b$, there is a basic
open set $V_1\in T([0,1/2])$ containing $1/2$
for which $\widehat{f}_1(V_1)\subseteq U$.
The only basic open sets $V_1\in T([0,1/2])$ containing $1/2$
are of the form $(s,1/2]$, so $V_1 = (s,1/2]$
for some $0 < s < 1/2$.
By a similar argument, there is a set $V_2 = [1/2,t)
\in T([1/2,1])$ with $1/2 < t < 1$ 
for which $\widehat{f}_2(V_2)\subseteq U$.
Now let $V = V_1\cup V_2 = (s,t)$. This set
is a basic open set in $T([0,1])$, and
$f(V) = \widehat{f}_1(V_1)\cup \widehat{f}_2(V_2)\subseteq U$.
Thus $f$ is continuous at $1/2$. \hfill $\Box$

\bigskip

\noindent
{\bf Corollary}. The relation $a\sim b$,
defined to hold if $a\stackrel{f}{\leadsto} b$
for some $f$, is an equivalence relation.

{\it Proof.} This relation is reflexive since
the constant function $f:[0,1]\rightarrow X: r\mapsto a$
is a path from $a$ to itself. 
This relation is symmetric since
if $f(r)$ is a path from $a$ to $b$, then 
$f(1-r)$ is a path from $b$ to $a$.
The theorem proves that the relation is 
transitive. \hfill $\Box$

\bigskip

The equivalence classes of the relation defined in the corollary
are called the {\it path components} of the space $X$.
Note that while the ordinary connected components
of a space must be closed, the path components
may be neither open nor closed.
\end{document}

