BRIEF ANSWERS TO TEST 1 (PDF file for Test 1)

1. • A collection B of subsets of X is a basis for a topology on X if
     • for all x in X there is U in B such that U contains x.
     • For all U, V in B and all x in (U intersect V) there is a W in B such that x is in W and W is a subset of (U intersect V).

   • Any collection S of subsets of X is a subbasis for a topology on X.

2. • The real numbers with the cofinite topology (other examples exist).

   • The real numbers with the cofinite topology (other examples exist).

3. • (--->) If A is clopen, then A and X-A are both closed. Therefore Bd(A) = (cl(A) intersect cl(X-A)) = (A intersect (X-A)) = emptyset.

   • (<---) If Bd(A) = emptyset, then cl(A) = (Bd(A) union int(A)) = int(A) is both open and closed. But since cl(A) contains A and int(A) is contained in A, it follows that cl(A) = A = int(A), so A is clopen.

4. To show that f(x)=x^2 is continuous with respect to the cofinite topology, it is enough to show that if C is a closed set then f^(-1)(C) is closed. If C = R, then f^(-1)(C) = R, which is closed. Otherwise C = {a1,...,an} is a finite set, and f^(-1)(C) = {+/-(a1)^(1/2), ..., +/-(an)^(1/2)}, which has at most twice as many points as C. In this case f^(-1)(C) is finite, hence closed.

BRIEF ANSWERS TO TEST 2 (PDF file for Test 2)

1. • False. If, for example, X is infinite and Y has size greater than 1, then for any y in Y the set X x {y} is an infinite proper closed set of X x Y in the product topology. But the cofinite topology on X x Y has no infinite proper closed set.

   • True. If X is connected, then so is the product X x X. Connectedness is a homeomorphism invariant, so Y x Y is connected. The continuous image of a connected space is connected, so by projecting onto a factor we get that the space Y is connected.

2. • A trivial space of more than one point (other examples exist). Here, any one point set is compact but not closed.

   • A trivial space of more than one point (other examples exist). Such a space is not T₃ since it is not T₁ (points are not closed). It is regular, since there do not exist a point p not contained in a nonempty closed set C (so there are no possible failures of regularity).

3. • (--->) A continuous image of the connected space X is connected, and the only nonempty connected subspaces of a discrete space are the singletons. Hence any continuous function of X into a discrete space is constant.
   • (<---) We argue the contrapositive. Assume that X is disconnected. Then X is the union of disjoint, nonempty, open sets U and V. The function from X to the discrete space {0,1} defined by f(x) = 0 if x is in U and f(x) = 1 is in V is nonconstant since U and V are nonempty, and is continuous since f ^-1(A) is one of {emptyset, U, V, X} (hence is open) for every subset A of {0,1}.

4. The set of fixed points of f is the equalizer of f and id, so it is closed.