Topologies, Subspaces, Closures, and Maps
Fall ’11 #topology/qual/completed
Let \(X\) be a topological space, and \(B \subset A \subset X\). Equip \(A\) with the subspace topology, and write \({ \operatorname{cl}}_X (B)\) or \({ \operatorname{cl}}_A (B)\) for the closure of \(B\) as a subset of, respectively, \(X\) or \(A\).
Determine, with proof, the general relationship between \({ \operatorname{cl}}_X (B) \cap A\) and \({ \operatorname{cl}}_A (B)\)
I.e., are they always equal? Is one always contained in the other but not conversely? Neither?
 Definition of closure: for \(A\subseteq X\), \({ \operatorname{cl}}_X(A)\) is the intersection of all \(B\supseteq A\) which are closed in \(X\).
 Definition of “relative” closure: for \(A\subseteq Y \subseteq X\), \(\operatorname{Cl}_Y(A)\) is the intersection of all \(B\) such that \(Y\supseteq B \supseteq A\) which are closed in \(Y\).
 Closed sets in a subspace: \(B' \subseteq Y\subseteq X\) is closed in \(Y\) if \(B' = B\cap Y\) for some \(B'\) closed in \(X\).
What’s the picture? Just need to remember what the closure with respect to a subspace looks like:
 Claim: \(\operatorname{Cl}_X(A) \cap Y = \operatorname{Cl}_Y(A)\).
 Write \(\operatorname{Cl}_Y(A)\) as the intersection of \(B'\) where \(Y\supseteq B' \supseteq A\) with \(B'\) closed in \(Y\).
 Every such \(B'\) is of the form \(B' = B \cap Y\) for some \(B\) closed in \(X\).
 Just identify the two sides directly by reindexing the intersection: \begin{align*} \operatorname{Cl}_Y(A) &\coloneqq\bigcap_{\substack{ Y\supseteq B' \supseteq A \\ B' \text{ closed in } Y}} B' \\ &= \bigcap_{\substack{ X \supseteq B \cap Y \supseteq A \\ B \text{ closed in } X}} \qty{ B \cap Y } \\ &= \qty{ \bigcap_{\substack{ X \supseteq B \cap Y \supseteq A \\ B \text{ closed in } X}} B} \cap Y \\ \\ &\coloneqq \operatorname{Cl}_X(A) \cap Y .\end{align*}
6 (Fall ’05) #topology/qual/completed
Prove that the unit interval \(I\) is compact. Be sure to explicitly state any properties of \({\mathbf{R}}\) that you use.
 Cantor’s intersection theorem: for a topological space, any nested sequence of compact nonempty sets has nonempty intersection.
 Bases for standard topology on \({\mathbf{R}}\).
 Definition of compactness
What’s the picture? Similar to covering \(\left\{{1\over n}\right\}\cup\left\{{0}\right\}\): cover \(x=0\) with one set, which nets all but finitely many points.
Proceed by contradiction. Binary search down into nested intervals, none of which have finite covers. Get a single point, a single set which eventually contains all small enough nested intervals. Only need finitely many more opens to cover the rest.
 Toward a contradiction, let \(\left\{{U_\alpha}\right\} \rightrightarrows[0, 1]\) be an open cover with no finite subcover.
 Then either \([0, {1\over 2}]\) or \([{1\over 2}, 1]\) has no finite subcover; WLOG assume it is \([0, {1\over 2}]\).
 Then either \([0, {1\over 4}]\) or \([{1\over 4}, {1\over 2}]\) has no finite subcover
 Inductively defining \([a_n, b_n]\) this way yields a sequence of compact nested intervals (each with no finite subcover) so Cantor’s Nested Interval theorem applies.
 Since \({\mathbf{R}}\) is a complete metric space and the diameters \({\operatorname{diam}}([a_n, b_n]) \leq {1 \over 2^n} \to 0\), the intersection contains exactly one point.
 Since \(p\in [0, 1]\) and the \(U_\alpha\) form an open cover, \(p\in U_\alpha\) for some \(\alpha\).
 Since a basis for \(\tau({\mathbf{R}})\) is given by open intervals, we can find an \({\varepsilon}>0\) such that \((p{\varepsilon}, p+{\varepsilon}) \subseteq U_\alpha\)
 Then if \({1\over 2^N} < {\varepsilon}\), for \(n\geq N\) we have \begin{align*}[a_n, b_n] \subseteq (p{\varepsilon}, p+{\varepsilon}) \subseteq U_\alpha.\end{align*}
 But then \(U_\alpha \rightrightarrows[a_n, b_n]\), yielding a finite subcover of \([a_n, b_n]\), a contradiction.
7 (Fall ’06). #topology/qual/work
A topological space is sequentially compact if every infinite sequence in \(X\) has a convergent subsequence.
Prove that every compact metric space is sequentially compact.
8 (Fall ’10). #topology/qual/completed
Show that for any two topological spaces \(X\) and \(Y\) , \(X \times Y\) is compact if and only if both \(X\) and \(Y\) are compact.
 Proof of the tube lemma:
 Continuous image of compact is compact.
What’s the picture?
Take an open cover of the product, use that vertical fibers are compact to get a finite cover for each fiber. Use tube lemma to get opens in the base space, run over all \(x\) so the tube bases cover \(X\). Use that \(X\) is compact to get a finite subcover.
\(\impliedby\):
 By the universal property, the product \(X\times Y\) is equipped with continuous projections \(\pi_X: X\times Y\to X\) and \(\pi_Y: X\times Y\to X\).
 The continuous image of a compact space is compact, and the images are all of \(X\) and \(Y\) respectively: \begin{align*} \pi_1(X\times Y) &= X \\ \pi_2(X\times Y) &= Y .\end{align*}
\(\implies\):

Let \(\left\{{U_j}\right\} \rightrightarrows X\times Y\) be an open cover.

Cover a fiber: fix \(x\in X\), the slice \(x \times Y\) is homeomorphic to \(Y\) and thus compact

Cover it by finitely many elements \(\left\{{U_j}\right\}_{j\leq m} \rightrightarrows{x} \times Y\).
Really, cover \(Y\), and then cross with \(x\) to cover \(x \times Y\).
 Set \begin{align*} N_x \coloneqq\bigcup_{j\leq m} U_j \supseteq x \times Y .\end{align*}

Apply the tube lemma to \(N_x\):
 Produce a neighborhood \(W_x\) of \(x\) in \(X\) where \(W_x \subset N_x\)
 This yields a finite cover: \begin{align*} \left\{{U_j}\right\}_{j\leq m}\rightrightarrows N_x \times Y \supset W_x \times Y \implies \left\{{U_j}\right\}_{j\leq m} \rightrightarrows W_x\times Y .\end{align*}

Cover the base: let \(x\in X\) vary: for each \(x\in X\), produce \(W_x \times Y\) as above, then \(\left\{{W_x}\right\}_{x\in X} \rightrightarrows X\) where each tube \(W_x \times Y\) is covered by finitely many \(U_j\).

Use that \(X\) is compact to produce a finite subcover \(\left\{{W_k}\right\}_{k \leq M} \rightrightarrows X\).

Then \(\left\{{W_k\times Y}\right\}_{k\leq M} \rightrightarrows X\times Y\), this is a finite set since each fiber was covered by finitely many opens
 Finitely many \(k\)
 For each \(k\), the tube \(W_k \times Y\) is covered by finitely by \(U_j\)
 And finite \(\times\) finite = finite.
12 (Spring ’06). #topology/qual/work
Write \(Y\) for the interval \([0, \infty)\), equipped with the usual topology.
Find, with proof, all subspaces \(Z\) of \(Y\) which are retracts of \(Y\).
\todo[inline]{Not finished. Add concepts}

Using the fact that \([0, \infty) \subset {\mathbf{R}}\) is Hausdorff, any retract must be closed, so any closed interval \([{\varepsilon}, N]\) for \(0\leq {\varepsilon}\leq N \leq \infty\).
 Note that \({\varepsilon}= N\) yields all one point sets \(\left\{{x_0}\right\}\) for \(x_0 \geq 0\).
 No finite discrete sets occur, since the retract of a connected set is connected.
13 (Fall ’06). #topology/qual/work

Prove that if the space \(X\) is connected and locally path connected then \(X\) is path connected.

Is the converse true? Prove or give a counterexample.
14 (Fall ’07) #topology/qual/work
Let \(\left\{{X_\alpha \mathrel{\Big}\alpha \in A}\right\}\) be a family of connected subspaces of a space \(X\) such that there is a point \(p \in X\) which is in each of the \(X_\alpha\).
Show that the union of the \(X_\alpha\) is connected.
\todo[inline]{Proof 2 not complete?}
 Take two connected sets \(X, Y\); then there exists \(p\in X\cap Y\).
 Toward a contradiction: write \(X\cup Y = A {\textstyle\coprod}B\) with both \(A, B \subset A{\textstyle\coprod}B\) open.
 Since \(p\in X \cup Y = A{\textstyle\coprod}B\), WLOG \(p\in A\). We will show \(B\) must be empty.

Claim: \(A\cap X\) is clopen in \(X\).
 \(A\cap X\) is open in \(X\): ?
 \(A\cap X\) is closed in \(X\): ?
 The only clopen sets of a connected set are empty or the entire thing, and since \(p\in A\), we must have \(A\cap X = X\).
 By the same argument, \(A\cap Y = Y\).
 So \(A\cap\qty{X\cup Y} = \qty{A\cap X} \cup\qty{A\cap Y} = X\cup Y\)
 Since \(A\subset X\cup Y\), \(A\cap\qty{X\cup Y} = A\)
 Thus \(A = X\cup Y\), forcing \(B = \emptyset\).
Let \(X \coloneqq\cup_\alpha X_\alpha\), and let \(p\in \cap X_\alpha\). Suppose toward a contradiction that \(X = A {\textstyle\coprod}B\) with \(A,B\) nonempty, disjoint, and relatively open as subspaces of \(X\). Wlog, suppose \(p\in A\), so let \(q\in B\) be arbitrary.
Then \(q\in X_\alpha\) for some \(\alpha\), so \(q\in B \cap X_\alpha\). We also have \(p\in A \cap X_\alpha\).
But then these two sets disconnect \(X_\alpha\), which was assumed to be connected – a contradiction.
5 (Fall ’04). #topology/qual/work
Let \(X\) be a topological space.

Prove that \(X\) is connected if and only if there is no continuous nonconstant map to the discrete twopoint space \(\left\{{0, 1}\right\}\).

Suppose in addition that \(X\) is compact and \(Y\) is a connected Hausdorff space. Suppose further that there is a continuous map \(f : X \to Y\) such that every preimage \(f^{1}(y)\) for \(y \in Y\), is a connected subset of \(X\).
Show that \(X\) is connected.
 Give an example showing that the conclusion of (b) may be false if \(X\) is not compact.
? (Spring ’10) #topology/qual/completed
If \(X\) is a topological space and \(S \subset X\), define in terms of open subsets of \(X\) what it means for \(S\) not to be connected.
Show that if \(S\) is not connected there are nonempty subsets \(A, B \subset X\) such that \begin{align*} A \cup B = S {\quad \operatorname{and} \quad} A \cap \overline{B} = \overline{A} \cap B = \emptyset \end{align*}
Here \(\overline{A}\) and \(\overline{B}\) denote closure with respect to the topology on the ambient space \(X\).

Topic: closure and connectedness in the subspace topology.
 See Munkres p.148
 Lemma: \(X\) is connected iff the only subsets of \(X\) that are closed and open are \(\emptyset, X\).

\(S\subset X\) is **not ** connected if \(S\) with the subspace topology is not connected.

I.e. there exist \(A, B \subset S\) such that
 \(A, B \neq \emptyset\),
 \(A\cap B = \emptyset\),
 \(A {\textstyle\coprod}B = S\).

I.e. there exist \(A, B \subset S\) such that
 Or equivalently, there exists a nontrivial \(A\subset S\) that is clopen in \(S\).
Show stronger statement: this is an iff.
\(\implies\):
 Suppose \(S\) is not connected; we then have sets \(A \cup B = S\) from above and it suffices to show \({ \operatorname{cl}}_Y(A) \cap B = A \cap{ \operatorname{cl}}_X(B) = \emptyset\).
 \(A\) is open by assumption and \(Y\setminus A = B\) is closed in \(Y\), so \(A\) is clopen.
 Write \({ \operatorname{cl}}_Y(A) \coloneqq{ \operatorname{cl}}_X(A) \cap Y\).
 Since \(A\) is closed in \(Y\), \(A = { \operatorname{cl}}_Y(A)\) by definition, so \(A = { \operatorname{cl}}_Y(A) = { \operatorname{cl}}_X(A) \cap Y\).
 Since \(A\cap B = \emptyset\), we then have \({ \operatorname{cl}}_Y(A) \cap B = \emptyset\).
 The same argument applies to \(B\), so \({ \operatorname{cl}}_Y(B) \cap A = \emptyset\).
\(\impliedby\):
 Suppose displayed condition holds; given such \(A, B\) we will show they are clopen in \(Y\).

Since \({ \operatorname{cl}}_Y(A) \cap B = \emptyset\), (claim) we have \({ \operatorname{cl}}_Y(A) = A\) and thus \(A\) is closed in \(Y\).

Why?
\begin{align*} { \operatorname{cl}}_Y(A) &\coloneqq{ \operatorname{cl}}_X(A) \cap Y \\ &= { \operatorname{cl}}_X(A) \cap\qty{A{\textstyle\coprod}B} \\ &= \qty{{ \operatorname{cl}}_X(A) \cap A} {\textstyle\coprod}\qty{{ \operatorname{cl}}_X(A) \cap B} \\ &= A {\textstyle\coprod}\qty{{ \operatorname{cl}}_X(A) \cap B} \quad\text{since } A \subset { \operatorname{cl}}_Y(A) \\ &= A {\textstyle\coprod}\qty{{ \operatorname{cl}}_Y(A) \cap B} \quad \text{since } B \subset Y \\ &= A {\textstyle\coprod}\emptyset \quad\text{using the assumption} \\ &= A .\end{align*}

Why?
 But \(A = Y\setminus B\) where \(B\) is closed, so \(A\) is open and thus a nontrivial clopen subset.
If \(S\subset X\) is not connected, then there exists a subset \(A\subset S\) that is both open and closed in the subspace topology, where \(A\neq \emptyset, S\).
Suppose \(S\) is not connected, then choose \(A\) as above. Then \(B = S\setminus A\) yields a pair \(A, B\) that disconnects \(S\). Since \(A\) is closed in \(S\), \(\overline{A} = A\) and thus \(\overline{A} \cap B = A \cap B = \emptyset\). Similarly, since \(A\) is open, \(B\) is closed, and \(\overline{B} = B \implies \overline{B} \cap A = B \cap A = \emptyset\).
? (Spring ’11) #topology/qual/work
A topological space is totally disconnected if its only connected subsets are onepoint sets.
Is it true that if \(X\) has the discrete topology, it is totally disconnected?
Is the converse true? Justify your answers.
21 (Fall ’14) #topology/qual/work
Let \(X\) and \(Y\) be topological spaces and let \(f : X \to Y\) be a function.
Suppose that \(X = A \cup B\) where \(A\) and \(B\) are closed subsets, and that the restrictions \(f \mathrel{\Big}_A\) and \(f \mathrel{\Big}_B\) are continuous (where \(A\) and \(B\) have the subspace topology).
Prove that \(f\) is continuous.
23 (Spring ’15) #topology/qual/completed
Define a family \({\mathcal{T}}\) of subsets of \({\mathbf{R}}\) by saying that \(A \in T\) is \(\iff A = \emptyset\) or \({\mathbf{R}}\setminus A\) is a finite set.
Prove that \({\mathcal{T}}\) is a topology on \({\mathbf{R}}\), and that \({\mathbf{R}}\) is compact with respect to this topology.
 This is precisely the cofinite topology.
 \({\mathbf{R}}\in \tau\) since \({\mathbf{R}}\setminus {\mathbf{R}}= \emptyset\) is trivially a finite set, and \(\emptyset \in \tau\) by definition.
 If \(U_i \in \tau\) then \((\cup_i U_i)^c = \cap U_i^c\) is an intersection of finite sets and thus finite, so \(\cup_i U_i \in \tau\).
 If \(U_i \in \tau\), then \((\cap_{i=1}^n U_i)^c = \cup_{i=1}^n U_i^c\) is a finite union of finite sets and thus finite, so \(\cap U_i \in \tau\).
So \(\tau\) forms a topology.
To see that \(({\mathbf{R}}, \tau)\) is compact, let \(\left\{{U_i}\right\} \rightrightarrows {\mathbf{R}}\) be an open cover by elements in \(\tau\).
Fix any \(U_\alpha\), then \(U_\alpha^c = \left\{{p_1, \cdots, p_n}\right\}\) is finite, say of size \(n\). So pick \(U_1 \ni p_1, \cdots, U_n \ni p_n\); then \({\mathbf{R}}\subset U_\alpha \cup_{i=1}^n U_i\) is a finite cover.
25 (Fall ’16) #topology/qual/work
Let \({\mathcal{S}}, {\mathcal{T}}\) be topologies on a set \(X\). Show that \({\mathcal{S}}\cap {\mathcal{T}}\) is a topology on \(X\).
Give an example to show that \({\mathcal{S}}\cup {\mathcal{T}}\) need not be a topology.
42 (Spring ’10) #topology/qual/completed
Define an equivalence relation \(\sim\) on \({\mathbf{R}}\) by \(x \sim y\) if and only if \(x  y \in {\mathbf{Q}}\). Let \(X\) be the set of equivalence classes, endowed with the quotient topology induced by the canonical projection \(\pi : {\mathbf{R}}\to X\).
Describe, with proof, all open subsets of \(X\) with respect to this topology.
43 (Fall ’12) #topology/qual/work
Let \(A\) denote a subset of points of \(S^2\) that looks exactly like the capital letter A. Let \(Q\) be the quotient of \(S^2\) given by identifying all points of \(A\) to a single point.
Show that \(Q\) is homeomorphic to a familiar topological space and identify that space.
Compactness and Metric Spaces
1 (Spring ’06) #topology/qual/work
Suppose \((X, d)\) is a metric space. State criteria for continuity of a function \(f : X \to X\) in terms of:

open sets;

\({\varepsilon}\)’s and \(\delta\)’s; and

convergent sequences.
Then prove that (iii) implies (i).
26 (Fall ’17) #topology/qual/work
Let \(f : X \to Y\) be a continuous function between topological spaces.
Let \(A\) be a subset of \(X\) and let \(f (A)\) be its image in \(Y\) .
One of the following statements is true and one is false. Decide which is which, prove the true statement, and provide a counterexample to the false statement:

If \(A\) is closed then \(f (A)\) is closed.

If \(A\) is compact then \(f (A)\) is compact.
2 (Spring ’12) #topology/qual/work
Let \(X\) be a topological space.

State what it means for \(X\) to be compact.

Let \(X = \left\{{0}\right\} \cup \left\{{{1\over n} \mathrel{\Big}n \in {\mathbf{Z}}^+ }\right\}\). Is \(X\) compact?
 Let \(X = (0, 1]\). Is \(X\) compact?
\todo[inline]{Incomplete proof for part 3.}
See Munkres p.164, especially for (ii).

See definitions in review doc.

Direct proof:
 Let \(\left\{{U_i {~\mathrel{\Big\vert}~}j\in J}\right\}\rightrightarrows X\); then \(0\in U_j\) for some \(j\in J\).

In the subspace topology, \(U_i\) is given by some \(V\in \tau({\mathbf{R}})\) such that \(V\cap X = U_i\)
 A basis for the subspace topology on \({\mathbf{R}}\) is open intervals, so write \(V\) as a union of open intervals \(V = \cup_{k\in K} I_k\).
 Since \(0\in U_j\), \(0\in I_k\) for some \(k\).
 Since \(I_k\) is an interval, it contains infinitely many points of the form \(x_n = {1 \over n} \in X\)
 Then \(I_k \cap X \subset U_j\) contains infinitely many such points.
 So there are only finitely many points in \(X\setminus U_j\), each of which is in \(U_{j(n)}\) for some \(j(n) \in J\) depending on \(n\).
 Todo
\todo[inline]{Need direct proof}
3 (Spring ’09) #topology/qual/work
Let \((X, d)\) be a compact metric space, and let \(f : X \to X\) be an isometry: \begin{align*} \forall~ x, y \in X, \qquad d(f (x), f (y)) = d(x, y) .\end{align*} Prove that \(f\) is a bijection.
4 (Spring ’05) #topology/qual/completed
Suppose \((X, d)\) is a compact metric space and \(U\) is an open covering of \(X\).
Prove that there is a number \(\delta > 0\) such that for every \(x \in X\), the ball of radius \(\delta\) centered at \(x\) is contained in some element of \(U\).
Statement: show that the Lebesgue number is welldefined for compact metric spaces.
Note: this is a question about the Lebesgue Number. See Wikipedia for detailed proof.
 Write \(U = \left\{{U_i {~\mathrel{\Big\vert}~}i\in I}\right\}\), then \(X \subseteq \cup_{i\in I} U_i\). Need to construct a \(\delta > 0\).
 By compactness of \(X\), choose a finite subcover \(U_1, \cdots, U_n\).

Define the distance between a point \(x\) and a set \(Y\subset X\): \(d(x, Y) = \inf_{y\in Y} d(x, y)\).
 Claim: the function \(d({}, Y): X\to {\mathbf{R}}\) is continuous for a fixed set.
 Proof: Todo, not obvious.

Define a function
\begin{align*} f: X &\to {\mathbf{R}}\\ x &\mapsto {1\over n} \sum_{i=1}^n d(x, X\setminus U_i) .\end{align*}
 Note this is a sum of continuous functions and thus continuous.

Claim:
\begin{align*}\delta \coloneqq\inf_{x\in X}f(x) = \min_{x\in X}f(x) = f(x_{\text{min}}) > 0\end{align*}
suffices.
 That the infimum is a minimum: \(f\) is a continuous function on a compact set, apply the extreme value theorem: it attains its minimum.

That \(\delta > 0\): otherwise, \(\delta = 0 \implies \exists x_0\) such that \(d(x_0, X\setminus U_i) = 0\) for all \(i\).
 Forces \(x_0 \in X\setminus U_i\) for all \(i\), but \(X\setminus \cup U_i = \emptyset\) since the \(U_i\) cover \(X\).

That it satisfies the Lebesgue condition:
\begin{align*}\forall x\in X, \exists i {\quad \operatorname{such that} \quad} B_\delta(x) \subset U_i\end{align*}
 Let \(B_\delta(x) \ni x\); then by minimality \(f(x) \geq \delta\).
 Thus it can not be the case that \(d(x, X\setminus U_i) < \delta\) for every \(i\), otherwise \begin{align*}f(x) \leq {1\over n}\qty{ \delta + \cdots + \delta} = {n\delta \over n} = \delta\end{align*}
 So there is some particular \(i\) such that \(d(x, X\setminus U_i) \geq \delta\).
 But then \(B_\delta \subseteq U_i\) as desired.
44 (Spring ’15) #topology/qual/work

Prove that a topological space that has a countable base for its topology also contains a countable dense subset.

Prove that the converse to (a) holds if the space is a metric space.
18 (Fall ’07) #topology/qual/completed
Prove that if \((X, d)\) is a compact metric space, \(f : X \to X\) is a continuous map, and \(C\) is a constant with \(0 < C < 1\) such that \begin{align*} d(f (x), f (y)) \leq C \cdot d(x, y) \quad \forall x, y ,\end{align*} then \(f\) has a fixed point.

Define a new function
\begin{align*} g: X \to {\mathbf{R}}\\ x &\mapsto d_X(x, f(x)) .\end{align*}

Attempt to minimize. Claim: \(g\) is a continuous function.

Given claim, a continuous function on a compact space attains its infimum, so set
\begin{align*} m \coloneqq\inf_{x\in X} g(x) \end{align*}
and produce \(x_0\in X\) such that \(g(x) = m\). 
Then
\begin{align*} m> 0 \iff d(x_0, f(x_0)) > 0 \iff x_0 \neq f(x_0) .\end{align*}

Now apply \(f\) and use the assumption that \(f\) is a contraction to contradict minimality of \(m\):
\begin{align*} d(f(f(x_0)), f(x_0)) &\leq C\cdot d(f(x_0), x_0) \\ &< d(f(x_0), x_0) \quad\text{since } C<1\\ &\leq m \end{align*}

Proof that \(g\) is continuous: use the definition of \(g\), the triangle inequality, and that \(f\) is a contraction:
\begin{align*} d(x, f(x)) &\leq d(x, y) + d(y, f(y)) + d(f(x), f(y)) \\ \implies d(x, f(x))  d(y, f(y)) &\leq d(x, y) + d(f(x), f(y)) \\ \implies g(x)  g(y) &\leq d(x, y) + C\cdot d(x, y) = (C+1) \cdot d(x, y)\\ \end{align*}
 This shows that \(g\) is Lipschitz continuous with constant \(C+1\) (implies uniformly continuous, but not used).
19 (Spring ’15) #topology/qual/completed
Prove that the product of two connected topological spaces is connected.

Use the fact that a union of spaces containing a common point is still connected.

Fix a point \((a, b) \in X \times Y\).

Since the horizontal slice \(X_b\coloneqq X \times\left\{{b}\right\}\) is homeomorphic to \(X\) which is connected, as are all of the vertical slices \(Y_x \coloneqq\left\{{x}\right\} \times Y \cong Y\) (for any \(x\)), the “Tshaped” space \(T_x \coloneqq X_b \cup Y_x\) is connected for each \(x\).

Note that \((a, b) \in T_x\) for every \(x\), so \(\cup_{x\in X} T_x = X \times Y\) is connected.
20 (Fall ’14) #topology/qual/completed
Define what it means for a topological space to be:

Connected

Locally connected
 Give, with proof, an example of a space that is connected but not locally connected.
\todo[inline]{What's the picture?}
 Consider \({\mathbf{R}}\), unions of intervals, \({\mathbf{Q}}\), and the topologists sine curve.
See definitions in review doc.
\(X\coloneqq\) the Topologist’s sine curve suffices.

Claim 1: \(X\) is connected.
 Intervals and graphs of cts functions are connected, so the only problem point is \(0\).

Claim 2: \(X\) is not locally connected.
 Take any \(B_{\varepsilon}(0) \in {\mathbf{R}}^2\); then projecting onto the subspace \(\pi_X(B_{\varepsilon}(0))\) yields infinitely many arcs, each intersecting the graph at two points on \({{\partial}}B_{\varepsilon}(0)\).
 These are homeomorphic to a collection of disjoint embedded open intervals, and any disjoint union of intervals is clearly not connected.
22 (Fall ’18) #topology/qual/work
Let \(X\) be a compact space and let \(f : X \times R \to R\) be a continuous function such that \(f (x, 0) > 0\) for all \(x \in X\).
Prove that there is \({\varepsilon}> 0\) such that \(f (x, t) > 0\) whenever \({\left\lvert {t} \right\rvert} < {\varepsilon}\).
Moreover give an example showing that this conclusion may not hold if \(X\) is not assumed compact.
24 (Spring ’16) #topology/qual/work
In each part of this problem \(X\) is a compact topological space. Give a proof or a counterexample for each statement.

If \(\left\{{F_n }\right\}_{n=1}^\infty\) is a sequence of nonempty closed subsets of \(X\) such that \(F_{n+1} \subset F_{n}\) for all \(n\) then \begin{align*}\cap^\infty_{n=1} F_n\neq \emptyset.\end{align*}

If \(\left\{{O_n}\right\}_{n=1}^\infty\) is a sequence of nonempty open subsets of \(X\) such that \(O_{n+1} \subset O_n\) for all \(n\) then \begin{align*}\cap_{n=1}^\infty O_{n}\neq \emptyset.\end{align*}
27 (Fall ’17) #topology/qual/work
A metric space is said to be totally bounded if for every \({\varepsilon}> 0\) there exists a finite cover of \(X\) by open balls of radius \({\varepsilon}\).

Show: a metric space \(X\) is totally bounded iff every sequence in \(X\) has a Cauchy subsequence.

Exhibit a complete metric space \(X\) and a closed subset \(A\) of \(X\) that is bounded but not totally bounded.
You are not required to prove that your example has the stated properties.
 Use diagonal trick to construct the Cauchy sequence.
\(\implies\):
If \(X\) is totally bounded, let \(\varepsilon = \frac 1 n\) for each \(n\), and let \(\left\{{x_i}\right\}\) be an arbitrary sequence. For \(n=1\), pick a finite open cover \(\left\{{U_i}\right\}_n\) such that \({\operatorname{diam}}{U_i} < \frac 1 n\) for every \(i\).
Choose \(V_1\) such that there are infinitely many \(x_i \in V_1\). (Why?) Note that \({\operatorname{diam}}V_i < 1\). Now choose \(x_i \in V_1\) arbitrarily and define it to be \(y_1\).
Then since \(V_1\) is totally bounded, repeat this process to obtain \(V_2 \subseteq V_1\) with \({\operatorname{diam}}(V_2)< \frac 1 2\), and choose \(x_i \in V_2\) arbitrarily and define it to be \(y_2\).
This yields a nested family of sets \(V_1 \supseteq V_2 \supseteq \cdots\) and a sequence \(\left\{{y_i}\right\}\) such that \(d(y_i, y_j) < \max(\frac 1 i, \frac 1 j) \to 0\), so \(\left\{{y_i}\right\}\) is a Cauchy subsequence.
\(\impliedby\):
Then fix \(\varepsilon > 0\) and pick \(x_1\) arbitrarily and define \(S_1 = B(\varepsilon, x_1)\). Then pick \(x_2 \in S_1^c\) and define \(S_2 = S_1 \cup B(\varepsilon, x_2)\), and so on. Continue by picking \(x_{n+1} \in S_n^c\) (Since \(X\) is not totally bounded, this can always be done) and defining \(S_{n+1} = S_n \cup B(\varepsilon, x_{n+1})\).
Then \(\left\{{x_n}\right\}\) is not Cauchy, because \(d(x_i, x_j) > \varepsilon\) for every \(i\neq j\).
Take \(X = C^0([0, 1])\) with the supnorm, then \(f_n(x) = x^n\) are all bounded by 1, but \({\left\lVert {f_i  f_j} \right\rVert} = 1\) for every \(i, j\), so no subsequence can be Cauchy, so \(X\) can not be totally bounded.
Moreover, \(\left\{{f_n}\right\}\) is closed. (Why?)
Spring ’19 #1 #topology/qual/completed
Is every complete bounded metric space compact? If so, give a proof; if not, give a counterexample.
\todo[inline]{Review, from last year.}

Complete and totally bounded \(\implies\) compact.

Definition: A space \(X\) is totally bounded if for every \(\varepsilon >0\), there is a finite cover \(X \subseteq \cup_\alpha B_\alpha(\varepsilon)\) such that the radius of each ball is less than \(\varepsilon\).

Definition: A subset of a space \(S \subset X\) is bounded if there exists a \(B(r)\) such that \(r<\infty\) and \(S \subseteq B(r)\)

Totally bounded \(\implies\) bounded
 Counterexample to converse: \({\mathbb{N}}\) with the discrete metric.
 Equivalent for Euclidean metric

Compact \(\implies\) totally bounded.

Counterexample for problem: the unit ball in any Hilbert (or Banach) space of infinite dimension is closed, bounded, and not compact.

Second counterexample: \(({\mathbf{R}}, (x,y) \mapsto \frac{{\left\lvert {xy} \right\rvert}}{1 + {\left\lvert {xy} \right\rvert}})\).

Best counterexample: \(X = \left({\mathbf{Z}}, ~\rho ( x , y ) = \left\{ \begin{array} { l l } { 1 } & { \text { if } x \neq y } \\ { 0 } & { \text { if } x = y } \end{array} \right.\right)\). This metric makes \(X\) complete for any \(X\), then take \({\mathbb{N}}\subset X\). All sets are closed, and bounded, so we have a complete, closed, bounded set that is not compact – take that cover \(U_i = B(1, i)\).

Useful tool: \((X, d) \cong_{\text{Top}} (X, \min{(d(x,y), 1)}\) where the RHS is now a bounded space. This preserves all topological properties (e.g. compactness).
Inductively, let \(\mathbf{x}_1 \in B(1, \mathbf{0})\) and \(A_1 = \mathop{\mathrm{span}}{(\mathbf{x}_1)}\), then choose \(s = \mathbf{x} + A_1 \in B(1,0)/A_1\) such that \({\left\lVert {s} \right\rVert} = \frac 1 2\) and then a representative \(\mathbf{x}_2\) such that \({\left\lVert {\mathbf{x}_2} \right\rVert} \leq 1\). Then \({\left\lVert {\mathbf{x}_2  \mathbf{x}_1} \right\rVert} \geq \frac 1 2\)
Then, let \(A_2 = \mathrm{span}(\mathbf{x}_1, \mathbf{x}_2)\), (which is closed) and repeat this for \(s = \mathbf{x} + A_2 \in B(1, \mathbf{0})/ A_2\) to get an \(\mathbf{x}_3\) such that \({\left\lVert {\mathbf{x}_3  \mathbf{x}_{\leq 2}} \right\rVert} \geq \frac 1 2\).
This produces a nonconvergent sequence in the closed ball, so it can not be compact.
Spring 2019 #2 #topology/qual/completed
Let \(X\) be Hausdorff, and recall that the onepoint compactification \(\tilde X\) is given by the following:

As a set, \(\tilde X \coloneqq X{\textstyle\coprod}\left\{{\infty}\right\}\).

A subset \(U\subseteq \tilde X\) is open iff either \(U\) is open in \(X\) or is of the form \(U = V{\textstyle\coprod}\left\{{\infty}\right\}\) where \(V\subset X\) is arbitrary and \(X\setminus V\) is compact.
Prove that this description defines a topology on \(\tilde X\) making \(\tilde X\) compact.
Definition: \((X, \tau)\) where \(\tau \subseteq \mathcal P(X)\) is a topological space iff
 \(\emptyset, X \in \tau\)
 \(\left\{{U_i}\right\}_{i\in I} \subseteq \tau \implies \cup_{i\in I} U_i \in \tau\)
 \(\left\{{U_i}\right\}_{i\in {\mathbb{N}}} \subseteq \tau \implies \cap_{i\in {\mathbb{N}}} U_i \in \tau\)
We can write \(\overline{(X, \tau)} = (X {\textstyle\coprod}{\operatorname{pt}}, \tau \cup\tau')\) where \(\tau' = \left\{{U{\textstyle\coprod}{\operatorname{pt}}{~\mathrel{\Big\vert}~}XU ~\text{is compact}}\right\}\). We need to show that \(T \coloneqq\tau \cup\tau'\) forms a topology.
 We have \(\emptyset,X \in \tau \implies \emptyset, X \in \tau \cup\tau'\).
 We just need to check that \(\tau'\) is closed under arbitrary unions. Let \(\left\{{U_i}\right\} \subset \tau'\), so \(XU_i = K_i\) a compact set for each \(i\). Then \(\cup_{i} U_i = \cup_i X (XU_i)= \cup_i X  K_i = X  \cup_i K_i\)
Spring 2021 #3 #topology/qual/work
For nonempty subsets \(A, B\) of a metric space \((X, d)\), define the setwise distance as \begin{align*} d(A, B) \coloneqq\inf \left\{{ d(a, b) {~\mathrel{\Big\vert}~}a\in A,\, b\in B }\right\} .\end{align*}

Suppose that \(A\) and \(B\) are compact. Show that there is an \(a\in A\) and \(b\in B\) such that \(d(A, B) = d(a, b)\).

Suppose that \(A\) is closed and \(B\) is compact. Show that if \(d(A, B) = 0\) then \(A \cap B = \emptyset\).
 Give an example in which \(A\) is closed, \(B\) is compact, and \(d(a, b) > d(A, B)\) for all \(a\in A\) and \(b\in B\).
Hint: take \(X = \left\{{ 0 }\right\} \cup(1, 2] \subset {\mathbf{R}}\). Throughout this problem, you may use without proof that the map \(d:X\times X\to {\mathbf{R}}\) is continuous.
Connectedness
9 (Spring ’13) #topology/qual/work
Recall that a topological space is said to be connected if there does not exist a pair \(U, V\) of disjoint nonempty subsets whose union is \(X\).

Prove that \(X\) is connected if and only if the only subsets of \(X\) that are both open and closed are \(X\) and the empty set.

Suppose that \(X\) is connected and let \(f : X \to {\mathbf{R}}\) be a continuous map. If \(a\) and \(b\) are two points of \(X\) and \(r\) is a point of \({\mathbf{R}}\) lying between \(f (a)\) and \(f (b)\) show that there exists a point \(c\) of \(X\) such that \(f (c) = r\).
10 (Fall ’05) #topology/qual/completed
Let \begin{align*} X = \left\{{(0, y) \mathrel{\Big} 1 \leq y \leq 1}\right\} \cup \left\{{\qty{x, s = \sin\qty{1 \over x}} \mathrel{\Big}0 < x \leq 1}\right\} .\end{align*}
Prove that \(X\) is connected but not path connected.
\(X\) is connected:
 Write \(X = L{\textstyle\coprod}G\) where \(L = \left\{{0}\right\} \times[1, 1]\) and \(G = \left\{{\Gamma(\sin(x)) {~\mathrel{\Big\vert}~}x\in (0, 1]}\right\}\) is the graph of \(\sin(x)\).

\(L \cong [0, 1]\) which is connected
 Claim: Every interval is connected (todo)

Claim: \(G\) is connected (i.e. as the graph of a continuous function on a connected set)

The function
\begin{align*} f: (0, 1] &\to [1, 1] \\ x &\mapsto \sin(x) \end{align*}
is continuous (how to prove?)  Products of continuous functions are continuous iff all of the components are continuous.

Claim: The diagonal map \(\Delta: Y\to Y\times Y\) where \(\Delta(t) = (t, t)\) is continuous for any \(Y\) since \(\Delta = (\operatorname{id}, \operatorname{id})\)
 Product of identity functions, which are continuous.

The composition of continuous function is continuous, therefore
\begin{align*} F : (0, 1] &\xrightarrow{\Delta} (0, 1]^2 \xrightarrow{(\operatorname{id}, f)} (0, 1] \times[1, 1] \\ t &\mapsto (t, t) \mapsto (t, f(t)) \end{align*}
 Then \(G = F((0, 1])\) is the continuous image of a connected set and thus connected.

The function

Claim: \(X\) is connected
 Suppose there is a disconnecting cover \(X = A{\textstyle\coprod}B\) such that \(\overline{A} \cap B = A\cap\overline{B} = \emptyset\) and \(A, B \neq \emptyset\).
 WLOG let \((x, \sin(x))\in B\) for \(x>0\) (otherwise just relabeling \(A, B\))

Claim: \(B = G\)
 It can’t be the case that \(A\) intersects \(G\): otherwise \begin{align*}X = A{\textstyle\coprod}B \implies G = (A\cap G) {\textstyle\coprod}(B \cap V)\end{align*} disconnects \(G\). So \(A\cap G = \emptyset\), forcing \(A \subseteq L\)
 Similarly \(L\) can not be disconnected, so \(B\cap L = \emptyset\) forcing \(B \subset G\)
 So \(A \subset L\) and \(B\subset G\), and since \(X = A{\textstyle\coprod}B\), this forces \(A = L\) and \(B = G\).

But any open set \(U\) in the subspace topology \(L\subset {\mathbf{R}}^2\) (generated by open balls) containing \((0, 0) \in L\) is the restriction of a ball \(V \subset {\mathbf{R}}^2\) of radius \(r>0\), i.e. \(U = V \cap X\).
 But any such ball contains points of \(G\): \begin{align*}n\gg 0 \implies {1 \over n\pi} < r \implies \exists g\in G \text{ s.t. } g\in U.\end{align*}
 So \(U \cap L \cap G \neq \emptyset\), contradicting \(L\cap G = \emptyset\).

Claim: \(X\) is not pathconnected.

Todo: “can’t get from \(L\) to \(G\) in finite time”.

Toward a contradiction, choose a continuous function \(f:I \to X\) with \(f(0) \in G\) and \(f(1) \in L\).
 Since \(L \cong [0, 1]\), use pathconnectedness to create a path \(f(1) \to (0, 1)\)
 Concatenate paths and reparameterize to obtain \(f(1) = (0, 1) \in L \subset {\mathbf{R}}^2\).

Let \({\varepsilon}= {1\over 2}\); by continuity there exists a \(\delta\in I\) such that \begin{align*} t\in B_\delta(1) \subset I \implies f(t) \in B_{\varepsilon}(\mathbf{0}) \in X \end{align*}

Using the fact that \([1\delta, 1]\) is connected, \(f([1\delta, 1]) \subset X\) is connected.

Let \(f(1\delta) = \mathbf{x}_0 = (x_0, y_0) \subset X\subset {\mathbf{R}}^2\).

Define a composite map
\begin{align*} F: [0, 1] &\to {\mathbf{R}} F &\coloneqq{\operatorname{pr}}_{x{\hbox{}}\text{axis}} \circ f .\end{align*}
 \(F\) is continuous as a composition of continuous functions.

Then \(F([1\delta, 1]) \subset {\mathbf{R}}\) is connected and thus must be an interval \((a, b)\)

Since \(f(1) = \mathbf{0}\) which has \(x{\hbox{}}\)component zero, \([0, b] \subset (a, b)\).

Since \(f(1\delta) = \mathbf{x}\), \(F(\mathbf{x}) = x_0\) and this \([0, x_0] \subset (a, b)\).

Thus for all \(x \in (0, x_0]\) there exists a \(t\in [1\delta, 1]\) such that \(f(t) = (x, \sin\qty{1\over x})\).

Now toward the contradiction, choose \(x = {1 \over 2n\pi  \pi/2} \in {\mathbf{R}}\) with \(n\) large enough such that \(x\in (0, x_0)\).
 Note that \(\sin\qty{1\over x} = 1\) by construction.
 Apply the previous statement: there exists a \(t\) such that \(f(t) = (x, \sin\qty{1\over x}) = (x, 1)\).
 But then \begin{align*} {\left\lVert {f(t)  f(x)} \right\rVert} = {\left\lVert {(x, 1)  (0, 1)} \right\rVert} = {\left\lVert {(x, 2)} \right\rVert} > {1\over 2} ,\end{align*} contradicting continuity of \(f\).

Let \(X = A \cup B\) with \(A = \left\{{(0, y) {~\mathrel{\Big\vert}~}y\in [1, 1] }\right\}\) and \(B = \left\{{(x, \sin(1/x)) {~\mathrel{\Big\vert}~}x\in (0, 1]}\right\}\). Since \(B\) is the graph of a continuous function, which is always connected. Moreover, \(X = \overline{A}\), and the closure of a connected set is still connected.
Alternative direct argument: the subspace \(X' = B \cup\left\{{\mathbf{0}}\right\}\) is not connected. If it were, write \(X' = U {\textstyle\coprod}V\), where wlog \(\mathbf{0} \in U\). Then there is an open such that \(\mathbf{0} \in N_r(\mathbf{0}) \subset U\). But any neighborhood about zero intersects \(B\), so we must have \(V \subset B\) as a strict inclusion. But then \(U \cap B\) and \(V\) disconnects \(B\), a connected set, which is a contradiction.
To see that \(X\) is not pathconnected, suppose toward a contradiction that there is a continuous function \(f: I \to X \subset {\mathbf{R}}^2\). In particular, \(f\) is continuous at \(\mathbf{0}\), and so
\begin{align*} \forall \varepsilon \quad \exists \delta {~\mathrel{\Big\vert}~}{\left\lVert {\mathbf{x}} \right\rVert} < \delta \implies {\left\lVert {f(\mathbf{x})} \right\rVert} < \varepsilon .\end{align*}
where the norm is the standard Euclidean norm.
However, we can pick \(\varepsilon < 1\), say, and consider points of the form \(\mathbf{x}_n = (\frac{1}{2n\pi}, 0)\). In particular, we can pick \(n\) large enough such that \({\left\lVert {\mathbf{x}_n} \right\rVert}\) is as small as we like, whereas \({\left\lVert {f(\mathbf{x}_n)} \right\rVert} = 1 > \varepsilon\) for all \(n\), a contradiction.
11 (Fall ’18) #topology/qual/work
Let
\begin{align*} X=\left\{(x, y) \in \mathbb{R}^{2}  x>0, y \geq 0, \text { and } \frac{y}{x} \text { is rational }\right\} \end{align*}and equip \(X\) with the subspace topology induced by the usual topology on \({\mathbf{R}}^2\).
Prove or disprove that \(X\) is connected.
\todo[inline]{Not convincing..}

Consider the (continuous) projection \(\pi: {\mathbf{R}}^2 \to {\mathbf{RP}}^1\) given by \((x, y) \mapsto [y/x, 1]\) in homogeneous coordinates.
 I.e. this sends points to lines through the origin with rational slope).

Note that the image of \(\pi\) is \({\mathbf{RP}}^1\setminus\left\{{\infty}\right\}\), which is homeomorphic to \({\mathbf{R}}\).

If we now define \(f = {\left.{{\pi}} \right_{{X}} }\), we have \(f(X) \twoheadrightarrow{\mathbf{Q}}\subset {\mathbf{R}}\).

If \(X\) were connected, then \(f(X)\) would also be connected, but \({\mathbf{Q}}\subset {\mathbf{R}}\) is disconnected, a contradiction.
Spring 2021 #2
Let \(X \coloneqq\prod_{==1}^{\infty} \left\{{ 0, 1 }\right\}\) endowed with the product topology.

Show that for all points \(x,y\in X\) with \(x\neq y\), there are open subsets \(U_x, U_y \subset X\) such that \(x\in U_x, y\in U_y\), with \(U_x \cup U_y = X\) and \(U_x \cap U_y = \emptyset\).

Show that \(X\) is totally disconnected, i.e. the only nonempty connected subsets of \(X\) are singletons.
Hausdorff Spaces and Separation
29 (Fall ’14) #topology/qual/work
Is every product (finite or infinite) of Hausdorff spaces Hausdorff? If yes, prove it. If no, give a counterexample.
30 (Spring ’18) #topology/qual/completed
Suppose that \(X\) is a Hausdorff topological space and that \(A \subset X\). Prove that if \(A\) is compact in the subspace topology then \(A\) is closed as a subset of X.

Let \(A \subset X\) be compact, and pick a fixed \(x\in X\setminus A\).

Since \(X\) is Hausdorff, for arbitrary \(a\in A\), there exists opens \(U_{a} \ni a\) and \(U_{x,a}\ni x\) such that \(V_{a} \cap U_{x,a} = \emptyset\).

Then \(\left\{{U_{a} {~\mathrel{\Big\vert}~}a\in A}\right\} \rightrightarrows A\), so by compactness there is a finite subcover \(\left\{{U_{a_i}}\right\} \rightrightarrows A\).

Now take \(U = \cup_i U_{a_i}\) and \(V_x = \cap_i V_{a_i, x}\), so \(U\cap V = \emptyset\).
 Note that both \(U\) and \(V_x\) are open.

But then defining \(V \coloneqq\cup_{x\in X\setminus A} V_x\), we have \(X\setminus A \subset V\) and \(V\cap A = \emptyset\), so \(V = X\setminus A\), which is open and thus \(A\) is closed.
31 (Spring ’09) #topology/qual/completed

Show that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism.

Give an example that shows that the “Hausdorff” hypothesis in part (a) is necessary.
 Continuous bijection + open map (or closed map) \(\implies\) homeomorphism.
 Closed subsets of compact sets are compact.
 The continuous image of a compact set is compact.
 Compact subsets of Hausdorff spaces are closed.
We’ll show that \(f\) is a closed map.
Let \(U \in X\) be closed.
 Since \(X\) is compact, \(U\) is compact
 Since \(f\) is continuous, \(f(U)\) is compact
 Since \(Y\) is Hausdorff, \(f(U)\) is closed.
Note that any finite space is clearly compact.
Take \(f: ([2], \tau_1) \to ([2], \tau_2)\) to be the identity map, where \(\tau_1\) is the discrete topology and \(\tau_2\) is the indiscrete topology. Any map into an indiscrete topology is continuous, and \(f\) is clearly a bijection.
Let \(g\) be the inverse map; then note that \(1 \in \tau_1\) but \(g^{1}(1) = 1\) is not in \(\tau_2\), so \(g\) is not continuous.
32 (Fall ’14) #topology/qual/completed
Let \(X\) be a topological space and let \begin{align*} \Delta = \left\{{(x, y) \in X \times X \mathrel{\Big}x = y}\right\} .\end{align*}
Show that \(X\) is a Hausdorff space if and only if \(\Delta\) is closed in \(X \times X\).
\(\implies\):
 Let \(p\in X^2\setminus \Delta\).
 Then \(p\) is of the form \((x, y)\) where \(x\neq y\) and \(x,y\in X\).
 Since \(X\) is Hausdorff, pick \(N_x, N_y\) in \(X\) such that \(N_x \cap N_y = \emptyset\).
 Then \(N_p\coloneqq N_x \times N_y\) is an open set in \(X^2\) containing \(p\).

Claim: \(N_p \cap\Delta = \emptyset\).
 If \(q \in N_p \cap\Delta\), then \(q = (z, z)\) where \(z\in X\), and \(q\in N_p \implies q\in N_x \cap N_y = \emptyset\).
 Then \(X^2\setminus \Delta = \cup_p N_p\) is open.
\(\impliedby\):
 Let \(x\neq y\in X\).
 Consider \((x, y) \in \Delta^c \subset X^2\), which is open.
 Thus \((x, y) \in B\) for some box in the product topology.
 \(B = U \times V\) where \(U\ni x, V\ni y\) are open in \(X\), and \(B \subset X^2\setminus \Delta\).

Claim: \(U\cap V = \emptyset\).
 Otherwise, \(z\in U\cap V \implies (z, z) \in B\cap\Delta\), but \(B \subset X^2\setminus \Delta \implies B \cap\Delta = \emptyset\).
33 (Fall ’06) #topology/qual/work
If \(f\) is a function from \(X\) to \(Y\) , consider the graph \begin{align*} G = \left\{{(x, y) \in X \times Y \mathrel{\Big}f (x) = y}\right\} .\end{align*}

Prove that if \(f\) is continuous and \(Y\) is Hausdorff, then \(G\) is a closed subset of \(X \times Y\).

Prove that if \(G\) is closed and \(Y\) is compact, then \(f\) is continuous.
34 (Fall ’04) #topology/qual/work
Let X be a noncompact locally compact Hausdorff space, with topology \({\mathcal{T}}\). Let \(\tilde X = X \cup \left\{{\infty}\right\}\) (\(X\) with one point adjoined), and consider the family \({\mathcal{B}}\) of subsets of \(\tilde X\) defined by \begin{align*} {\mathcal{B}}= T \cup \left\{{S \cup \left\{{\infty}\right\}\mathrel{\Big}S \subset X,~~ X \backslash S \text{ is compact}}\right\} .\end{align*}

Prove that \({\mathcal{B}}\) is a topology on \(\tilde X\), that the resulting space is compact, and that \(X\) is dense in \(\tilde X\).

Prove that if \(Y \supset X\) is a compact space such that \(X\) is dense in \(Y\) and \(Y \backslash X\) is a singleton, then Y is homeomorphic to \(\tilde X\).
The space \(\tilde X\) is called the onepoint compactification of \(X\).

Find familiar spaces that are homeomorphic to the one point compactifications of

\(X = (0, 1)\) and
 \(X = {\mathbf{R}}^2\).
35 (Fall ’16) #topology/qual/work
Prove that a metric space \(X\) is normal, i.e. if \(A, B \subset X\) are closed and disjoint then there exist open sets \(A \subset U \subset X, ~B \subset V \subset X\) such that \(U \cap V = \emptyset\).
36 (Spring ’06) #topology/qual/work
Prove that every compact, Hausdorff topological space is normal.
37 (Spring ’09) #topology/qual/work
Show that a connected, normal topological space with more than a single point is uncountable.
38 (Spring ’08) #topology/qual/completed
Give an example of a quotient map in which the domain is Hausdorff, but the quotient is not.
 \({\mathbf{R}}\) is clearly Hausdorff, and \({\mathbf{R}}/{\mathbf{Q}}\) has the indiscrete topology, and is thus nonHausdorff.
 So take the quotient map \(\pi:{\mathbf{R}}\to {\mathbf{R}}/{\mathbf{Q}}\).
Direct proof that \({\mathbf{R}}/{\mathbf{Q}}\) isn’t Hausdorff:
 Pick \([x] \subset U \neq [y] \subset V \in {\mathbf{R}}/{\mathbf{Q}}\) and suppose \(U\cap V = \emptyset\).
 Pull back \(U\to A, V\to B\) open disjoint sets in \({\mathbf{R}}\)
 Both \(A, B\) contain intervals, so they contain rationals \(p\in A, q\in B\)
 Then \([p] = [q] \in U\cap V\).
39 (Fall ’04) #topology/qual/work
Let \(X\) be a compact Hausdorff space and suppose \(R \subset X \times X\) is a closed equivalence relation. Show that the quotient space \(X/R\) is Hausdorff.
40 (Spring ’18) #topology/qual/work
Let \(U \subset {\mathbf{R}}^n\) be an open set which is bounded in the standard Euclidean metric. Prove that the quotient space \({\mathbf{R}}^n / U\) is not Hausdorff.
41 (Fall ’09) #topology/qual/work
Let \(A\) be a closed subset of a normal topological space \(X\). Show that both \(A\) and the quotient \(X/A\) are normal.
45 (Spring ’11) #topology/qual/work
Recall that a topological space is regular if for every point \(p \in X\) and for every closed subset \(F \subset X\) not containing \(p\), there exist disjoint open sets \(U, V \subset X\) with \(p \in U\) and \(F \subset V\).
Let \(X\) be a regular space that has a countable basis for its topology, and let \(U\) be an open subset of \(X\).

Show that \(U\) is a countable union of closed subsets of \(X\).

Show that there is a continuous function \(f : X \to [0,1]\) such that \(f (x) > 0\) for \(x \in U\) and \(f (x) = 0\) for \(x \in U\).
Exercises
Basics
Exercise #topology/qual/work
Show that for \(A\subseteq X\), \({ \operatorname{cl}}_X(A)\) is the smallest closed subset containing \(A\).
Exercise #topology/qual/completed
Give an example of spaces \(A\subseteq B \subseteq X\) such that \(A\) is open in \(B\) but \(A\) is not open in \(X\).
\hfill
\hfill
No: Take \([0, 1] \subset [0, 1] \subset {\mathbf{R}}\). Then \([0, 1]\) is tautologically open in \([0, 1]\) as it is the entire space, But \([0, 1]\) is not open in \({\mathbf{R}}\):
 E.g. \(\left\{{1}\right\}\) is not an interior point (every neighborhood intersects the complement \({\mathbf{R}}\setminus[0, 1]\)).
Exercise #topology/qual/work
Show that the diagonal map \(\Delta(x) = (x, x)\) is continuous.
Exercise #topology/qual/work
Show that if \(A_i \subseteq X\), then \({ \operatorname{cl}}_X(\cup_i A_i) = \cup_i { \operatorname{cl}}_X(A_i)\).
Exercise #topology/qual/work
Show that \({\mathbf{R}}\) is not homeomorphic to \([0, \infty)\).
Exercise #topology/qual/work
Show that the set \((x, y) \in {\mathbf{R}}^2\) such that at least one of \(x, y\) is rational with the subspace topology is a connected space.
Exercise: \({\mathbf{R}}/{\mathbf{Q}}\) is indiscrete
Show that \({\mathbf{R}}/{\mathbf{Q}}\) has the indiscrete topology.
 Let \(U \subset {\mathbf{R}}/{\mathbf{Q}}\) be open and nonempty, show \(U = {\mathbf{R}}/{\mathbf{Q}}\).
 Let \([x] \in U\), then \(x \in \pi^{1}(U) \coloneqq V \subset{\mathbf{R}}\) is open.
 Then \(V\) contains an interval \((a, b)\)
 Every \(y\in V\) satisfies \(y+q \in V\) for all \(q\in {\mathbf{Q}}\), since \(y+qy \in {\mathbf{Q}}\implies [y+q] = [y]\).
 So \((aq, b+q) \in V\) for all \(q\in {\mathbf{Q}}\).
 So \(\cup_{q\in {\mathbf{Q}}}(aq, b+q) \in V \implies {\mathbf{R}}\subset V\).
 So \(\pi(V) = {\mathbf{R}}/{\mathbf{Q}}= U\), and thus the only open sets are the entire space and the empty set.
Connectedness
Exercise #topology/qual/work
Prove that \(X\) is connected iff the only clopen subsets are \(\emptyset, X\).
Exercise #topology/qual/work
Let \(A \subset X\) be a connected subspace.
Show that if \(B\subset X\) satisfies \(A\subseteq B \subseteq \overline{A}\), then \(B\) is connected.
Exercise #topology/qual/work
Show that:
 Connected does not imply path connected
 Connected and locally path connected does imply path connected
 Path connected implies connected
Exercise #topology/qual/work
Use the fact that intervals are connected to prove the intermediate value theorem.
Exercise #topology/qual/work
Prove that the continuous image of a connected set is connected.
Exercise #topology/qual/work
Show that if \(X\) is locally path connected, then
 Every open subset of \(X\) is again locally pathconnected.
 \(X\) is connected \(\iff X\) is pathconnected.
 Every path component of \(X\) is a connected component of \(X\).
 Every connected component of \(X\) is open in \(X\).
Exercise #topology/qual/completed
Show that \([0, 1]\) is connected.
\hfill
\hfill [Reference](https://sites.math.washington.edu/~morrow/334_16/connected.pdf) A potentially shorter proof
Let \(I = [0, 1] = A\cup B\) be a disconnection, so
 \(A, B \neq \emptyset\)
 \(A {\textstyle\coprod}B = I\)
 \({ \operatorname{cl}}_I(A) \cap B = A \cap{ \operatorname{cl}}_I(B) = \emptyset\). Let \(a\in A\) and \(b\in B\) where WLOG \(a<b\)
 (since either \(a<b\) or \(b<a\), and \(a\neq b\) since \(A, B\) are disjoint) Let \(K = [a, b]\) and define \(A_K \coloneqq A\cap K\) and \(B_K \coloneqq B\cap K\). Now \(A_K, B_K\) is a disconnection of \(K\). Let \(s = \sup(A_K)\), which exists since \({\mathbf{R}}\) is complete and has the LUB property Claim: \(s \in { \operatorname{cl}}_I(A_K)\). Proof:
 If \(s\in A_K\) there’s nothing to show since \(A_K \subset { \operatorname{cl}}_I(A_K)\), so assume \(s\in I\setminus A_K\).
 Now let \(N_s\) be an arbitrary neighborhood of \(s\), then using ??? we can find an \({\varepsilon}>0\) such that \(B_{\varepsilon}(s) \subset N_s\)
 Since \(s\) is a supremum, there exists an \(a\in A_K\) such that \(s{\varepsilon}< a\).
 But then \(a \in B_{\varepsilon}(s)\) and \(a\in N_s\) with \(a\neq s\).
 Since \(N_s\) was arbitrary, every \(N_s\) contains a point of \(A_K\) not equal to \(s\), so \(s\) is a limit point by definition. Since \(s\in { \operatorname{cl}}_I(A_K)\) and \({ \operatorname{cl}}_I(A_K)\cap B_K = \emptyset\), we have \(s\not \in B_K\). Then the subinterval \((x, b] \cap A_K = \emptyset\) for every \(x>c\) since \(c \coloneqq\sup A_K\). But since \(A_K {\textstyle\coprod}B_K = K\), we must have \((x, b] \subset B_K\), and thus \(s\in { \operatorname{cl}}_I(B_K)\). Since \(A_K, B_K\) were assumed disconnecting, \(s\not \in A_K\) But then \(s\in K\) but \(s\not\in A_K {\textstyle\coprod}B_K = K\), a contradiction.
Compactness
\(\star\) Exercise #topology/qual/completed
Let \(X\) be a compact space and let \(A\) be a closed subspace. Show that \(A\) is compact.
\hfill
\hfill
Let \(X\) be compact, \(A\subset X\) closed, and \(\left\{{U_\alpha}\right\} \rightrightarrows A\) be an open cover. By definition of the subspace topology, each \(U_\alpha = V_\alpha \cap A\) for some open \(V_\alpha \subset X\), and \(A\subset \cup_\alpha V_\alpha\). Since \(A\) is closed in \(X\), \(X\setminus A\) is open. Then \(\left\{{V_\alpha}\right\}\cup\left\{{X\setminus A}\right\}\rightrightarrows X\) is an open cover, since every point is either in \(A\) or \(X\setminus A\). By compactness of \(X\), there is a finite subcover \(\left\{{U_j {~\mathrel{\Big\vert}~}j\leq N}\right\}\cup\left\{{X\setminus A}\right\}\) Then \(\qty{\left\{{U_j}\right\} \cup\left\{{X\setminus A}\right\}} \cap A \coloneqq\left\{{V_j}\right\}\) is a finite cover of \(A\).
\(\star\) Exercise #topology/qual/completed
Let \(f : X \to Y\) be a continuous function, with \(X\) compact. Show that \(f(X)\) is compact.
\hfill
\hfill
Let \(f:X\to Y\) be continuous with \(X\) compact, and \(\left\{{U_\alpha}\right\} \rightrightarrows f(X)\) be an open cover. Then \(\left\{{f^{1}(U_\alpha)}\right\} \rightrightarrows X\) is an open cover of \(X\), since \(x\in X \implies f(x) \in f(X) \implies f(x) \in U_\alpha\) for some \(\alpha\), so \(x\in f^{1}(U_\alpha)\) by definition. By compactness of \(X\) there is a finite subcover \(\left\{{f^{1}(U_j) {~\mathrel{\Big\vert}~}j\leq N}\right\} \rightrightarrows X\). Then the finite subcover \(\left\{{U_j{~\mathrel{\Big\vert}~}j\leq N}\right\} \rightrightarrows f(X)\), since if \(y\in f(X)\), \(y\in U_\alpha\) for some \(\alpha\) and thus \(f^{1}(y) \in f^{1}(U_j)\) for some \(j\) since \(\left\{{U_j}\right\}\) is a cover of \(X\).
Let \(A\) be a compact subspace of a Hausdorff space \(X\). Show that \(A\) is closed.
Exercise #topology/qual/work
Show that any infinite set with the cofinite topology is compact.
Exercise #topology/qual/work
Show that every compact metric space is complete.
Exercise #topology/qual/work
Show that if \(X\) is second countable and Hausdorff, or a metric space, then TFAE:
 \(X\) is compact
 Every infinite subset \(A\subseteq X\) has a limit point in \(X\).
 Every sequence in \(X\) has a convergent subsequence in \(X\).
Exercise #topology/qual/work
Show that if \(f: A\to B\) is a continuous map between metric spaces and \(K\subset A\) is compact, then \({\left.{{f}} \right_{{K}} }\) is uniformly continuous.
Exercise #topology/qual/work
Show that if \(f:X\to Y\) is continuous and \(X\) is compact then \(f(X)\) is compact.
Exercise #topology/qual/work
Show that if \(f:X\to {\mathbf{R}}\) and \(X\) is compact then \(f\) is bounded and attains its min/max.
Exercise #topology/qual/work
Show that a finite product or union compact spaces is again compact.
Exercise #topology/qual/work
Show that a quotient of a compact space is again compact.
Exercise #topology/qual/work
Show that if \(X\) is compact and \(A\subseteq X\) is closed then \(A\) is compact.
Exercise #topology/qual/work
Show that if \(X\) is Hausdorff and \(A\subseteq X\) is compact then \(A\) is closed.
Exercise #topology/qual/work
Show that if \(X\) is a metric space and \(A\subseteq X\) is compact then \(A\) is bounded.
Exercise #topology/qual/work
Show that a continuous map from a compact space to a Hausdorff space is closed.
Exercise #topology/qual/work
Show that an injective continuous map from a compact space to a Hausdorff space is an embedding (a homeomorphism onto its image).
Exercise #topology/qual/work
Show that \([0, 1]\) is compact.
Exercise #topology/qual/work
Show that a compact Hausdorff space is is metrizable iff it is secondcountable.
Exercise #topology/qual/work
Show that if \(X\) is metrizable, then \(X\) is compact
Exercise #topology/qual/work
Give an example of a space that is compact but not sequentially compact, and vice versa.
Exercise #topology/qual/work
Show that a sequentially compact space is totally bounded.
Exercise #topology/qual/work
Show that \({\mathbf{R}}\) with the cofinite topology is compact.
Exercise #topology/qual/work
Show that \([0, 1]\) is compact without using the HeineBorel theorem.
Separation
Exercise #topology/qual/work
Show that \(X\) is Hausdorff iff \(\Delta(X)\) is closed in \(X\times X\).
Exercise #topology/qual/work
Prove that \(X, Y\) are Hausdorff iff \(X\times Y\) is Hausdorff.
Exercise #topology/qual/work
Show that \({\mathbf{R}}\) is separable.
Exercise #topology/qual/work
Show that any space with the indiscrete topology is separable.
Exercise #topology/qual/work
Show that any countable space with the discrete topology is separable.
Exercise #topology/qual/work
Show that the minimal uncountable order with the order topology is not separable.
Exercise #topology/qual/work
Show that every first countable space is second countable.
Exercise #topology/qual/work
Show that every metric space is Hausdorff in its metric topology.
Hausdorff Spaces
Exercise
Show that a compact set in a Hausdorff space is closed.
Exercise #topology/qual/completed
Let \(A\subset X\) with \(A\) closed and \(X\) compact, and show that \(A\) is compact.
Alternative definition of “open”: todo.
 Let \(A\) be a compact subset of \(X\) a Hausdorff space, we will show \(X\setminus A\) is open
 Fix \(x\in X\setminus A\).
 Since \(X\) is Hausdorff, for every \(y\in A\) we can find \(U_y \ni y\) and \(V_x(y) \ni x\) depending on \(y\) such that \(U_x(y) \cap U_y = \emptyset\).
 Then \(\left\{{U_y {~\mathrel{\Big\vert}~}y\in A}\right\} \rightrightarrows A\), and by compactness of \(A\) there is a finite subcover corresponding to a finite collection \(\left\{{y_1, \cdots, y_n}\right\}\).

Magic Step: set \(U = \cup U_{y_i}\) and \(V = \cap V_x(y_i)\);
 Note \(A\subset U\) and \(x\in V\)
 Note \(U\cap V = \emptyset\).
 Done: for every \(x\in X\setminus A\), we have found an open set \(V\ni x\) such that \(V\cap A = \emptyset\), so \(x\) is an interior point and a set is open iff every point is an interior point.
Exercise #topology/qual/completed
Show that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism.
 It suffices to show that \(f\) is a closed map, i.e. if \(U\subseteq X\) is closed then \(f(U)\subseteq Y\) is again closed.

Let \(U\in X\) be closed; since \(X\) is closed, \(U\) is compact
 Since closed subsets of compact spaces are compact.

Since \(f\) is continuous, \(f(U)\) is compact
 Since the continuous image of a compact set is compact.

Since \(Y\) is Hausdorff and \(f(U)\) is compact, \(f(U)\) is closed
 Since compact subsets of Hausdorff spaces are closed.
Exercise #topology/qual/work
Show that a closed subset of a Hausdorff space need not be compact.
Exercise #topology/qual/work
Show that in a compact Hausdorff space, \(A\) is closed iff \(A\) is compact.
Exercise #topology/qual/work
Show that a local homeomorphism between compact Hausdorff spaces is a covering space.