General Topology

• 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: $$\begin{align*} \operatorname{Cl}_Y(A) &\coloneqq\bigcap_{\substack{ Y\supseteq B' \supseteq A \\ B' \text{ closed in } Y}} B' \\ &= \bigcap_{\substack{ X \supseteq B \cap Y \supseteq A \\ B \text{ closed in } X}} \qty{ B \cap Y } \\ &= \qty{ \bigcap_{\substack{ X \supseteq B \cap Y \supseteq A \\ B \text{ closed in } X}} B} \cap Y \\ \\ &\coloneqq \operatorname{Cl}_X(A) \cap Y .\end{align*}$$

• 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.

• 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.

• ${\mathbf{R}}\in \tau$ 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.

• 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}

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.

• 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).

• 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.

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 $\cup_{i} U_i = \cup_i X- (X-U_i)= \cup_i X - K_i = X - \cup_i K_i$

• 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.

• 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.

$\implies$:

• 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 $B \subset X^2\setminus \Delta \implies B \cap\Delta = \emptyset$.

• ${\mathbf{R}}$ 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 $[p] = [q] \in U\cap V$.

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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 ${\mathbf{R}}\setminus[0, 1]$).

• 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.

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[Reference]

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.

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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 $A$.

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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 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.

• 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.