# General Topology

## 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 two-point 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
• $$A, B \neq \emptyset$$,
• $$A\cap B = \emptyset$$,
• $$A {\textstyle\coprod}B = S$$.
• 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*}
• 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 one-point sets.

Is it true that if $$X$$ has the discrete topology, it is totally disconnected?

### 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 well-defined 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 “T-shaped” 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 sup-norm, 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: $$({\mathbf{R}}, (x,y) \mapsto \frac{{\left\lvert {x-y} \right\rvert}}{1 + {\left\lvert {x-y} \right\rvert}})$$.

• 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 non-convergent 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 one-point 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 $$\tau' = \left\{{U{\textstyle\coprod}{\operatorname{pt}}{~\mathrel{\Big\vert}~}X-U ~\text{is compact}}\right\}$$. 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 $$X-U_i = K_i$$ a compact set for each $$i$$. Then $$\cup_{i} U_i = \cup_i X- (X-U_i)= \cup_i X - K_i = X - \cup_i K_i$$

### 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.
• 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 path-connected.
• 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 path-connectedness 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 path-connected, 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 one-point 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 non-Hausdorff.
• 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 $$y+q-y \in {\mathbf{Q}}\implies [y+q] = [y]$$.
• So $$(a-q, b+q) \in V$$ for all $$q\in {\mathbf{Q}}$$.
• So $$\cup_{q\in {\mathbf{Q}}}(a-q, b+q) \in V \implies {\mathbf{R}}\subset V$$.
• 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 path-connected.
• $$X$$ is connected $$\iff X$$ is path-connected.
• 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.

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[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.

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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.

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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 second-countable.

#### 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 Heine-Borel 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.

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