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.