General Topology Mastery Examination

A Cumulative Examination in General Topology

A useful topology proof rarely depends on one definition in isolation. Compactness turns an open cover into finite data; connectedness rules out certain separations; continuity transfers topological information between spaces; and separation axioms provide neighborhoods that distinguish points or closed sets. This examination brings these ideas together in problems where choosing the right theorem matters as much as carrying out the details.

The solutions use results established earlier in this course, including the Tube Lemma, the compactness of finite products, the continuous-image theorems for compact and connected spaces, and the shrinking characterization of normality. Two additional tools are proved here: a pasting lemma for maps defined on a finite closed cover, and Urysohn’s lemma, which constructs a continuous real-valued function separating disjoint closed sets in a normal space.

Problem 1: Pasting Continuous Maps

Suppose a space is covered by finitely many closed pieces, and a continuous map is specified on each piece. If the specifications agree wherever pieces overlap, they define one map on the whole space. The point requiring proof is continuity: continuity on each piece alone does not generally settle the matter for an arbitrary cover, but a finite closed cover gives a direct test using closed sets.

Proof. Because the \(F_i\) cover \(X\), at least one index \(i\) satisfies \(x\in F_i\) for each \(x\in X\). If \(x\) belongs to both \(F_i\) and \(F_j\), the agreement assumption gives \(f_i(x)=f_j(x)\). Thus the rule defining \(f(x)\) is independent of the choice of index, so \(f\) is well-defined.

Let \(C\) be any closed subset of \(Y\). For each \(i\), continuity of \(f_i\) implies that \(f_i^{-1}(C)\) is closed in the subspace \(F_i\). By the closed-set characterization of the subspace topology, there is a closed set \(D_i\subseteq X\) such that \(f_i^{-1}(C)=F_i\cap D_i\). Since \(F_i\) is closed in \(X\), this preimage is closed in \(X\). The definition of \(f\) now gives

This is a finite union of closed subsets of \(X\), hence is closed. The closed-set criterion for continuity therefore shows that \(f\) is continuous. \(\square\)

Problem 2: Compactness and Connectedness Together

Consider the map \(g:[0,1]\to\mathbb{R}^2\) defined by \(g(t)=(t,t^2)\). Determine whether its image is compact and connected, and whether it is closed in \(\mathbb{R}^2\).

A common exam error is to infer that a continuous image of a compact space is automatically closed in every target. The compact-image theorem gives compactness in the image’s subspace topology. To conclude closedness in the ambient target, one also needs a hypothesis such as Hausdorffness.

Problem 3: Separating Closed Sets by a Continuous Function

Normality gives a stronger form of separation than disjoint open neighborhoods: in a normal space, two disjoint closed sets can be assigned the distinct values \(0\) and \(1\) by a continuous function into \([0,1]\). The construction inserts open sets at dyadic rational levels and ensures that their closures remain nested. This is Urysohn’s lemma.

Proof. Write \(D=\{k/2^n: n\geq0,\ 0\leq k\leq2^n\}\) for the dyadic rationals in \([0,1]\). Set \(U_1=X\setminus B\). Because \(A\) is closed and \(A\subseteq U_1\), the shrinking characterization of normality gives an open set \(U_0\) such that

We construct open sets \(U_r\) for \(r\in D\), preserving \(\overline{U_r}\subseteq U_s\) whenever \(r<s\). Suppose sets have been chosen at one dyadic level, with this nesting property for consecutive levels. For each pair \(r<s\) of consecutive numbers already chosen, the closed set \(\overline{U_r}\) is contained in the open set \(U_s\). By the shrinking characterization, choose an open set \(U_{(r+s)/2}\) satisfying

Doing this for every consecutive pair at each successive dyadic level constructs all the required sets. The nesting property for any \(r<s\) in \(D\) follows by chaining the inclusions for the finitely many dyadic levels between them.

For \(x\in X\), define

The set whose infimum is taken is nonempty because it contains \(1\), and it is bounded below by \(0\). Thus \(f(x)\in[0,1]\). If \(a\in A\), then \(a\in U_0\), so \(f(a)=0\). If \(b\in B\), then \(b\notin U_1=X\setminus B\). Since \(U_r\subseteq U_1\) for \(r<1\), the point \(b\) belongs to none of the \(U_r\), and therefore \(f(b)=1\).

It remains to prove continuity. For a real number \(c\) with \(0<c\leq1\), the definition of the infimum gives

For the forward inclusion, if \(f(x)<c\), some element of the set defining the infimum is less than \(c\); that element cannot be \(1\), so it is an \(r\in D\) with \(x\in U_r\). For the reverse inclusion, \(x\in U_r\) with \(r<c\) implies \(f(x)\leq r<c\). Thus this sublevel set is open.

For \(0\leq c<1\), we also have

To prove the inclusion from right to left, suppose \(x\notin\overline{U_r}\). If \(x\in U_s\) for some \(s<r\), nesting would give \(x\in U_s\subseteq U_r\subseteq\overline{U_r}\), a contradiction. Thus every dyadic index at which \(x\) belongs to a set \(U_s\) is at least \(r\), and \(f(x)\geq r>c\). For the other inclusion, suppose \(f(x)>c\). Choose dyadic numbers \(r,s\) with \(c<r<s<f(x)\); this is possible because the dyadic rationals are dense in \([0,1]\). If \(x\in\overline{U_r}\), the nesting property gives \(\overline{U_r}\subseteq U_s\), so \(x\in U_s\), which implies \(f(x)\leq s\). This contradicts \(s<f(x)\). Hence \(x\notin\overline{U_r}\), proving the inclusion.

The displayed unions are open. For thresholds outside these ranges, the corresponding sublevel or superlevel sets are either empty or all of \(X\), since \(f\) takes values in \([0,1]\). Therefore the preimages of open rays are open, which proves that \(f:X\to[0,1]\) is continuous. \(\square\)

How to Use the Results in a Proof

The three problems illustrate a practical order of attack. First identify the conclusion: continuity, compactness, connectedness, closedness, or separation. Then check the hypotheses of a theorem that gives that conclusion. In the curve example, continuity is established before compactness and connectedness are transferred. Closedness is handled afterward, with Hausdorffness supplying the needed additional condition. For piecewise maps, the closed-cover pasting lemma is appropriate only after both agreement on overlaps and continuity on each piece have been checked.

Urysohn’s construction demonstrates a different proof technique: replace one difficult global choice by a countable sequence of compatible choices. At each stage, normality supplies the next open set with its closure still inside the required larger open set. The closure inclusions are not decorative; they are what make the later sublevel and superlevel sets open, and hence what makes the final function continuous.