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Bokuan Li
2026-03-06 14:06:15 -05:00
parent 173727665b
commit 5034bc4220
109 changed files with 1184 additions and 410 deletions
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5
src/topology/main/baire.tex
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@@ -36,6 +36,7 @@
\[
U \cap \bigcap_{n \in \natp}\ol V_n \subset U \cap \bigcap_{n \in \natp}U_n
\]
Since $\bigcap_{n \in \natp}\ol V_n \ne \emptyset$ by assumption (b), $U \cap \bigcap_{n \in \natp}U_n \ne \emptyset$, so $\bigcap_{n \in \natp}U_n$ is dense.
\end{proof}
@@ -48,11 +49,11 @@
\end{enumerate}
\end{theorem}
\begin{proof}
Let $U \subset X$ be open and $\seq{U_n} \subset 2^X$ be open and dense. Let $V_0 = U$. For each $n \in \natp$, by density of $U_n$, there exists $x \in U_n \cap V_{n - 1}$. Since $X$ is regular (\ref{definition:uniform-separated}/\ref{proposition:compact-hausdorff-normal}), there exists $V_{n} \in \cn^o(x)$ such that $x \in V_{n} \subset \ol U_{n} \subset U_n \cap V_{n-1}$. If $X$ is locally compact, choose $V_n$ to be precompact. If $X$ is completely metrisable, choose $V_n$ such that $\text{diam}(V_n) \le 1/n$.
Let $U \subset X$ be open and $\seq{U_n} \subset 2^X$ be open and dense. Let $V_0 = U$. For each $n \in \natp$, by density of $U_n$, there exists $x \in U_n \cap V_{n - 1}$. Since $X$ is regular (\autoref{definition:uniform-separated}/\autoref{proposition:compact-hausdorff-normal}), there exists $V_{n} \in \cn^o(x)$ such that $x \in V_{n} \subset \ol U_{n} \subset U_n \cap V_{n-1}$. If $X$ is locally compact, choose $V_n$ to be precompact. If $X$ is completely metrisable, choose $V_n$ such that $\text{diam}(V_n) \le 1/n$.
Now, if $X$ is locally compact, then by the finite intersection property, $\bigcap_{n \in \natp}\ol{V_n} \ne \emptyset$. If $X$ is completely metrisable, then $\seq{V_n}$ is a Cauchy filter base, and converges to at least one point, so $\bigcap_{n \in \natp}\ol{V_n} \ne \emptyset$.
By \ref{lemma:baire-condition}, $X$ is a Baire space.
By \autoref{lemma:baire-condition}, $X$ is a Baire space.
\end{proof}
\begin{definition}[Meagre]

6
src/topology/main/compact.tex
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@@ -17,26 +17,30 @@
\[
U_J = \bigcup_{j \in J}E_j^c
\]
then $U_J \subset X$ is open. For any $J, J' \subset I$, $U_J \cup U_{J'} = U_{J \cup J'}$.
Suppose for contradiction that $\bigcap_{i \in I}E_i = \emptyset$, then
\[
\mathbf{U} = \bracs{U_J|J \subset I \text{ finite}}
\]
is an open cover for $X$. By assumption, $U_J \subsetneq X$ for all $J \subset I$ finite. Thus $\mathbf{U}$ admits no finite subcover, contradiction.
(2) $\Rightarrow$ (3): Let $\fF \subset 2^X$ be a filter, then $\bracsn{\overline{E}| E \in \fF}$ satisfies the hypothesis of (2).
(3) $\Leftrightarrow$ (4): By \ref{definition:accumulation-point}, the cluster points and the limit points of an ultrafilter coincide.
(3) $\Leftrightarrow$ (4): By \autoref{definition:accumulation-point}, the cluster points and the limit points of an ultrafilter coincide.
(3) $\Rightarrow$ (1): For each $J \subset I$, let
\[
E_J = \bigcap_{j \in J}U_j^c
\]
then for each $J, J' \subset I$, $E_J \cap E_{J'} = E_{J \cup J'}$. Assume for contradiction that $\mathbf{U}$ admits no finite subcover. Let
\[
\fB = \bracs{E_H|J \subset I \text{ finite}}
\]
then $\fB$ is a filter base consisting of closed sets. By assumption, there exists $x \in \bigcap_{i \in I}U_j^c$, so $\mathbf{U}$ is not an open cover, contradiction.
\end{proof}

2
src/topology/main/connected.tex
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@@ -42,5 +42,5 @@
Let $X$ be a topological space and $A \subset X$ be connected, then there exists a unique connected set $C \supset A$ such that for any $C' \supset A$ connected, $C \supset C'$. The set $C$ is the \textbf{connected component} of $A$.
\end{definition}
\begin{proof}
Let $C$ be the union of all connected sets that contain $A$, then $C$ is connected by \ref{proposition:connected-union}, and is the maximum connected set containing $A$ by definition.
Let $C$ be the union of all connected sets that contain $A$, then $C$ is connected by \autoref{proposition:connected-union}, and is the maximum connected set containing $A$ by definition.
\end{proof}

4
src/topology/main/continuity.tex
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@@ -26,7 +26,7 @@
Local continuity, $(2) \Rightarrow (1)$: Since $f(\cn(x))$ converges to $f(x)$, for each $U \in \cn(f(x))$, there exists $V \in \cn(x)$ such that $f(V) \subset V$. Thus $V \subset f^{-1}(U) \in \cn(x)$ by (F1).
Global continuity, $(1) \Rightarrow (2)$: Let $U \subset Y$ be open, then $U \in \cn(f(x))$ for all $x \in f^{-1}(U)$. Hence $f^{-1}(U) \in \cn(x)$ for all $x \in f^{-1}(U)$. Thus $U$ is open by \ref{lemma:openneighbourhood}.
Global continuity, $(1) \Rightarrow (2)$: Let $U \subset Y$ be open, then $U \in \cn(f(x))$ for all $x \in f^{-1}(U)$. Hence $f^{-1}(U) \in \cn(x)$ for all $x \in f^{-1}(U)$. Thus $U$ is open by \autoref{lemma:openneighbourhood}.
Global continuity, $(2) \Rightarrow (1)$: Let $x \in X$ and $V \in \cn(f(x))$, then there exists $U \in \cn(f(x))$ open, so $f^{-1}(V) \supset f^{-1}(U)$ contains an open set that contains $x$. Therefore $f^{-1}(V) \in \cn(x)$.
@@ -43,7 +43,7 @@
then there exists a unique $f \in C(X; Y)$ such that $f|_{U_i} = f_i$ for all $i \in I$.
\end{lemma}
\begin{proof}
By \ref{lemma:glue-function}, there exists a unique $f: X \to Y$ with $f|_{U_i} = f_i$ for all $i \in I$.
By \autoref{lemma:glue-function}, there exists a unique $f: X \to Y$ with $f|_{U_i} = f_i$ for all $i \in I$.
Let $x \in X$. By assumption (a), there exists $i \in I$ such that $x \in U_i$. Since $f_i \in C(U_i; Y)$, for any $V \in \cn_Y(f(x))$, there exists $U \in \cn_{U_i}(x)$ such that $f_i(U) \subset V$. As $U_i$ is open in $X$, $U \in \cn_X(x)$. Thus $f(U) = f_i(U) \subset V$, and $f$ is continuous at $x$.
\end{proof}

11
src/topology/main/definition.tex
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@@ -51,6 +51,8 @@
\[
\topo = \topo(\cb) = \bracs{\bigcup_{i \in I}U_i \bigg | \seqi{U} \subset \cb, I \text{ index set}}
\]
Conversely, if $\cb \subset 2^X$ is a family such that:
\begin{enumerate}
\item[(TB1)] For every $x \in X$, there exists $U \in \cb$ such that $x \in U$.
@@ -65,6 +67,8 @@
\[
\paren{\bigcup_{i \in I}U_i} \cap \paren{\bigcup_{j \in J}V_j} = \bigcup_{i \in I}\bigcup_{j \in J}U_i \cap V_j
\]
Thus to show that $\topo(\cb)$ satisfies (O2), it is sufficient to show that $U \cap V \in \topo(\cb)$ for all $U, V \in \cb$. To this end, for every $x \in U \cap V$, there exists $W_x \in \cb$ such that $x \in W_x \subset U \cap V$. Therefore $U \cap V = \bigcup_{x \in U \cap V}W_x \in \topo(\cb)$, and $\topo(\cb)$ satisfies (O2).
By definition, $\topo(\cb)$ satisfies (O3).
@@ -76,14 +80,16 @@
\[
\topo(\ce) = \bracs{\bigcup_{i \in I}U_i \bigg | U_i \in \cb(\ce)}
\]
where
\[
\cb(\ce) = \bracs{\bigcap_{j = 1}^n U_j \bigg | \seqf{U_j} \subset \ce, n \in \nat^+}
\]
is a base for $\topo(\ce)$. The topology $\topo(\ce)$ is known as the topology \textbf{generated by} $\ce$.
\end{definition}
\begin{proof}
Since $\cb(\ce) \supset \ce$, $\cb(\ce)$ satisfies (TB1). In addition, $\cb(\ce)$ is closed under finite intersections, so it satisfies (TB2). By \ref{definition:base}, $\cb(\ce)$ is a base for $\topo(\ce)$.
Since $\cb(\ce) \supset \ce$, $\cb(\ce)$ satisfies (TB1). In addition, $\cb(\ce)$ is closed under finite intersections, so it satisfies (TB2). By \autoref{definition:base}, $\cb(\ce)$ is a base for $\topo(\ce)$.
\end{proof}
\begin{definition}[Initial Topology]
@@ -98,6 +104,7 @@
\mathcal{B} = \bracs{\bigcap_{j \in J}f_j^{-1}(U_j) \bigg | J \subset I \text{ finite}, U_j \in \topo_j}
\]
is a base for $\topo$.
\end{enumerate}
The topology $\topo$ is known a the \textbf{initial/weak topology} generated by the maps $\seqi{f}$.
@@ -107,6 +114,6 @@
\begin{enumerate}
\item For each $i \in I$, $\topo \supset \bracs{f_i^{-1}(U)|U \in \topo_i}$, so $f_i \in C(\topo; Y_i)$.
\item If $\mathcal{S}$ is a topology such that $f_i \in C(X, \mathcal{S}; Y_i)$, then $\bracs{f_i^{-1}(U)|U \in \topo_i} \subset \mathcal{S}$. Thus $\ce \subset \mathcal{S}$ and $\mathcal{S} \supset \topo$.
\item By \ref{definition:generated-topology}, $\cb$ is a base for $\topo$.
\item By \autoref{definition:generated-topology}, $\cb$ is a base for $\topo$.
\end{enumerate}
\end{proof}

15
src/topology/main/filters.tex
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@@ -30,6 +30,7 @@
\[
\fF = \bracs{F \subset X| \exists E \in \fB: E \subset F}
\]
\end{proposition}
\begin{proof}
Filter Base $\Rightarrow$ (FB1): Let $E, F \in \fB$, then $E \cap F \in \fF$. Thus there exists $G \in \fB$ such that $G \subset E \cap F$.
@@ -79,15 +80,17 @@ The smallest filter $\fF(\fB_0) \subset 2^X$ containing $\fB_0$ is the filter \t
\[
\fB = \bracs{\bigcap_{j = 1}^n E_j \bigg | \seqf{E_j} \subset \fB_0, n \in \nat^+}
\]
\end{definition}
\begin{proof}
For any $\seqf{E_j}, \bracsn{F_j}_1^m \subset \fB_0$,
\[
G = \paren{\bigcap_{j = 1}^n E_j} \cap \paren{\bigcap_{j = 1}^m F_j} \in \fB
\]
Thus $\fB$ satisfies (FB1). Since $\bigcap_{j = 1}^n E_j \ne \emptyset$, $\emptyset \not\in \fB$, and $\fB$ satisfies (FB2).
By \ref{proposition:filterbasecriterion}, $\fB$ is a base for the filter $\fF = \bracs{E \subset X| \exists F \in \fB: F \subset E}$.
By \autoref{proposition:filterbasecriterion}, $\fB$ is a base for the filter $\fF = \bracs{E \subset X| \exists F \in \fB: F \subset E}$.
If $\fF' \supset \fB_0$ is a filter, then $\fF \supset \fB$ by (F2), and $\fF' \supset \fF$ by (F1). Thus $\fF$ is the smallest filtetr containing $\fB$.
\end{proof}
@@ -97,7 +100,7 @@ The smallest filter $\fF(\fB_0) \subset 2^X$ containing $\fB_0$ is the filter \t
Let $X$ be a set, $\fF, \mathfrak{G} \subset 2^X$ be filter subbases. If for any $E \in \fF$ and $F \in \mathfrak{G}$, $E \cap F \ne \emptyset$, then there exists a filter $\fU \supset \fF \cup \mathfrak{G}$.
\end{lemma}
\begin{proof}
Let $\seqf{E_j} \subset \fF$ and $\seqf[m]{F_j} \subset \mathfrak{G}$, then $\bigcap_{j = 1}^n E_j \in \fF$ and $\bigcap_{j = 1}^m F_j \in \mathfrak{G}$. Thus their intersection is non-empty, and $\fF \cup \mathfrak{G}$ is a filter subbase. By \ref{definition:generatedfilter}, there exists a filter containing $\fF \cup \mathfrak{G}$.
Let $\seqf{E_j} \subset \fF$ and $\seqf[m]{F_j} \subset \mathfrak{G}$, then $\bigcap_{j = 1}^n E_j \in \fF$ and $\bigcap_{j = 1}^m F_j \in \mathfrak{G}$. Thus their intersection is non-empty, and $\fF \cup \mathfrak{G}$ is a filter subbase. By \autoref{definition:generatedfilter}, there exists a filter containing $\fF \cup \mathfrak{G}$.
\end{proof}
\begin{definition}[Ultrafilter]
@@ -111,7 +114,7 @@ The smallest filter $\fF(\fB_0) \subset 2^X$ containing $\fB_0$ is the filter \t
If the above holds, then $\fF$ is an \textbf{ultrafilter}.
\end{definition}
\begin{proof}
(1) $\Rightarrow$ (2): Let $E \subset X$ with $E^c \not\in \fF$, then $E \cap F \ne \emptyset$ for all $F \in \fF$. By \ref{lemma:filter-extend}, there exists a filter $\fU \supset \fF \cup \bracs{E}$. Since $\fF$ is maximal, $\bracs{E} \in \fU \subset \fF$.
(1) $\Rightarrow$ (2): Let $E \subset X$ with $E^c \not\in \fF$, then $E \cap F \ne \emptyset$ for all $F \in \fF$. By \autoref{lemma:filter-extend}, there exists a filter $\fU \supset \fF \cup \bracs{E}$. Since $\fF$ is maximal, $\bracs{E} \in \fU \subset \fF$.
(2) $\Rightarrow$ (1): Let $E \subset X$ with $E \not \in \fF$, then $E^c \in \fF$. Since $E \cap E^c = \emptyset$, there exists no filter containing $\fF \cup \bracs{E}$. Thus $\fF$ is maximal.
@@ -133,6 +136,7 @@ The smallest filter $\fF(\fB_0) \subset 2^X$ containing $\fB_0$ is the filter \t
\[
\mathbf{U} = \bracs{\fU \supset \fF| \fU \subset 2^X \text{ is a filter}}
\]
and order it by inclusion. For any chain $\mathbf{C} \subset \mathbf{U}$, let $\fB = \bigcup_{\fF \in \textbf{C}}\fF$, then $\fB$ is a filter subbase, and the filter $\fU \in \mathbf{U}$ generated by $\fB$ is an upper bound for $\mathbf{C}$. By Zorn's lemma, $\textbf{U}$ has a maximal element.
(2): Let $E \subset X$. If there exists $F \in \fF$ with $F \cap E^c = \emptyset$, then $E \supset F$ and $E \in \fF$. Otherwise, $\fF \cup \bracs{E^c}$ is a filter subbase, and there exists an ultrafilter containing it by (1). In which case, there exists an ultrafilter $\fU \supset \fF$ with $E \not\in \fU$.
@@ -152,7 +156,7 @@ The smallest filter $\fF(\fB_0) \subset 2^X$ containing $\fB_0$ is the filter \t
If $A \subset X$ and $\fB \subset 2^A$, then $\fB$ \textbf{converges} to $x$ if $\fF(\fB) \supset \bracsn{U \cap A| U \in \cn(x)}$.
\end{definition}
\begin{proof}
(1) $\Leftrightarrow$ (2): By (2) of \ref{lemma:ultrafilter}.
(1) $\Leftrightarrow$ (2): By (2) of \autoref{lemma:ultrafilter}.
(1) $\Rightarrow$ (3): $\cn(x)$ is a fundamental system of neighbourhoods at $x$.
@@ -173,7 +177,7 @@ The smallest filter $\fF(\fB_0) \subset 2^X$ containing $\fB_0$ is the filter \t
\begin{proof}
(1) $\Rightarrow$ (3): Let $U \in \cn(x)$, then $U \cap E \ne \emptyset$ for all $E \in \fB$.
(3) $\Rightarrow$ (4): By \ref{lemma:filter-extend}, there exists a filter $\fU \supset \cb(x) \cup \fB$. Since $\cb(x)$ and $\fB$ are bases for $\cn(x)$ and $\fF$, respectively, $\fU \supset \cn(x) \cup \fF$.
(3) $\Rightarrow$ (4): By \autoref{lemma:filter-extend}, there exists a filter $\fU \supset \cb(x) \cup \fB$. Since $\cb(x)$ and $\fB$ are bases for $\cn(x)$ and $\fF$, respectively, $\fU \supset \cn(x) \cup \fF$.
(4) $\Rightarrow$ (1): Let $U \in \cn(x)$, then since $\fU \supset \fB \cup \cn(x)$, $U \cap E \ne \emptyset$ for all $E \in \fB \subset \fF \subset \fU$. Thus $x \in \bigcap_{E \in \fF}\ol{E}$.
\end{proof}
@@ -187,5 +191,6 @@ The smallest filter $\fF(\fB_0) \subset 2^X$ containing $\fB_0$ is the filter \t
\[
y = \lim_{x \to x_0 \\ x \in A}f(x)
\]
is a \textbf{limit} of $f$ at $y$ with respect to $A$. If $A = X$, then $x \in A$ may be omitted.
\end{definition}

7
src/topology/main/hausdorff.tex
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@@ -16,17 +16,18 @@
If the above holds, then $X$ is a \textbf{T2/Hausdorff} space.
\end{definition}
\begin{proof}
$(1) \Rightarrow (2)$: Let $y \in X \setminus \bracs{x}$, then there exists $U \in \cn(x)$ and $V \in \cn(y)$ such that $U \cap V = \emptyset$. By (2) of \ref{definition:closure}, $y \not\in \overline{U} \subset \bigcap_{U \in \cn(x)}\overline{U}$.
$(1) \Rightarrow (2)$: Let $y \in X \setminus \bracs{x}$, then there exists $U \in \cn(x)$ and $V \in \cn(y)$ such that $U \cap V = \emptyset$. By (2) of \autoref{definition:closure}, $y \not\in \overline{U} \subset \bigcap_{U \in \cn(x)}\overline{U}$.
$(2) \Rightarrow (3)$: Let $\fF \subset 2^X$ be a filter and $x \in X$ such that $\cn(x) \subset \fF$, then
\[
\bracs{x} = \bigcap_{U \in \cn(x)}\ol{U} \supset \bigcap_{U \in \cn(x)}\ol{U} \supset \bracs{x}
\]
so $x$ is the only cluster point of $\fF$.
$(3) \Rightarrow (4)$: Let $\fF \subset 2^X$ be a filter. If $\fF$ converges to $x \in X$, then $x$ is a cluster point of $\fF$. As $\fF$ admits only one cluster point, $x$ is the only limit point of $\fF$.
$(4) \Rightarrow (5)$: Let $x \in \overline{\Delta}$, then by (4) of \ref{definition:closure}, there exists $\fF \subset 2^\Delta$ converging to $x$. Let $i, j \in I$, then for any $y \in \Delta$, $\pi_i(y) = \pi_j(y)$. Thus $\pi_i(\fF) = \pi_j(\fF)$. By \ref{proposition:productfilterconvergence}, $\pi_i(\fF)$ converges to $\pi_i(x)$ and $\pi_j(\fF)$ converges to $\pi_j(x)$. By assumption, $\pi_i(x) = \pi_j(x)$. Since this holds for all pairs $i, j \in I$, $x \in \Delta$.
$(4) \Rightarrow (5)$: Let $x \in \overline{\Delta}$, then by (4) of \autoref{definition:closure}, there exists $\fF \subset 2^\Delta$ converging to $x$. Let $i, j \in I$, then for any $y \in \Delta$, $\pi_i(y) = \pi_j(y)$. Thus $\pi_i(\fF) = \pi_j(\fF)$. By \autoref{proposition:productfilterconvergence}, $\pi_i(\fF)$ converges to $\pi_i(x)$ and $\pi_j(\fF)$ converges to $\pi_j(x)$. By assumption, $\pi_i(x) = \pi_j(x)$. Since this holds for all pairs $i, j \in I$, $x \in \Delta$.
$(5) \Rightarrow (6)$: Take $I = \bracs{1, 2}$.
@@ -38,7 +39,7 @@
Let $X$ be a topological space, $Y$ be a Hausdorff space, $A \subset X$ be a dense subset, and $F, G \in C(X; Y)$. If $F|_A = G|_A$, then $F = G$.
\end{proposition}
\begin{proof}
Let $x \in X$. By (4) \ref{definition:closure}, there exists a filter base $\fB \subset 2^A$ that converges to $x$. By (2) of local continuity, $F(\fB)$ converges to $F(x)$ and $G(\fB)$ converges to $G(x)$. Since $F(\fB) = G(\fB)$, $F(x) = G(x)$ by (4) of \ref{definition:hausdorff}.
Let $x \in X$. By (4) \autoref{definition:closure}, there exists a filter base $\fB \subset 2^A$ that converges to $x$. By (2) of local continuity, $F(\fB)$ converges to $F(x)$ and $G(\fB)$ converges to $G(x)$. Since $F(\fB) = G(\fB)$, $F(x) = G(x)$ by (4) of \autoref{definition:hausdorff}.
\end{proof}
\begin{proposition}

42
src/topology/main/index.tex
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@@ -1,24 +1,24 @@
\chapter{Topological Spaces}
\label{chap:topological-spaces}
\input{./src/topology/main/definition.tex}
\input{./src/topology/main/filters.tex}
\input{./src/topology/main/nets.tex}
\input{./src/topology/main/neighbourhoods.tex}
\input{./src/topology/main/interiorclosureboundary.tex}
\input{./src/topology/main/continuity.tex}
\input{./src/topology/main/product.tex}
\input{./src/topology/main/hausdorff.tex}
\input{./src/topology/main/regular.tex}
\input{./src/topology/main/normal.tex}
\input{./src/topology/main/quotient.tex}
\input{./src/topology/main/connected.tex}
\input{./src/topology/main/path-connected.tex}
\input{./src/topology/main/local-path-connected.tex}
\input{./src/topology/main/unity.tex}
\input{./src/topology/main/compact.tex}
\input{./src/topology/main/sigma-compact.tex}
\input{./src/topology/main/para.tex}
\input{./src/topology/main/support.tex}
\input{./src/topology/main/lch.tex}
\input{./src/topology/main/baire.tex}
\input{./definition.tex}
\input{./filters.tex}
\input{./nets.tex}
\input{./neighbourhoods.tex}
\input{./interiorclosureboundary.tex}
\input{./continuity.tex}
\input{./product.tex}
\input{./hausdorff.tex}
\input{./regular.tex}
\input{./normal.tex}
\input{./quotient.tex}
\input{./connected.tex}
\input{./path-connected.tex}
\input{./local-path-connected.tex}
\input{./unity.tex}
\input{./compact.tex}
\input{./sigma-compact.tex}
\input{./para.tex}
\input{./support.tex}
\input{./lch.tex}
\input{./baire.tex}

16
src/topology/main/interiorclosureboundary.tex
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@@ -18,7 +18,7 @@
$(3) \Rightarrow (1)$: By (F1) of $\cn(x)$, $A \in \cn(x)$.
Let $U \subset A$ be open, then $U \in \cn(x)$ for all $x \in U$ by \ref{lemma:openneighbourhood}. By (3), $U \subset A^o$, so $A^o$ contains every open subset of $A$. On the other hand, for any $x \in A^o$, (2) implies that there exists $U \in \cn^o(x)$ with $U \subset A$. In which case, $U \subset A^o$, and $A^o \in \cn(x)$. Therefore $A^o$ itself is open by \ref{lemma:openneighbourhood}.
Let $U \subset A$ be open, then $U \in \cn(x)$ for all $x \in U$ by \autoref{lemma:openneighbourhood}. By (3), $U \subset A^o$, so $A^o$ contains every open subset of $A$. On the other hand, for any $x \in A^o$, (2) implies that there exists $U \in \cn^o(x)$ with $U \subset A$. In which case, $U \subset A^o$, and $A^o \in \cn(x)$. Therefore $A^o$ itself is open by \autoref{lemma:openneighbourhood}.
\end{proof}
\begin{definition}[Closure]
@@ -72,9 +72,9 @@
If the above holds, then $A$ is a \textbf{dense} subset of $X$.
\end{definition}
\begin{proof}
$(1) \Rightarrow (2)$: Since $U \ne \emptyset$, there exists $x \in U$, so $U \in \cn(x)$. By (2) of \ref{definition:closure}, $U \cap A \ne \emptyset$.
$(1) \Rightarrow (2)$: Since $U \ne \emptyset$, there exists $x \in U$, so $U \in \cn(x)$. By (2) of \autoref{definition:closure}, $U \cap A \ne \emptyset$.
$(2) \Rightarrow (3)$: Let $\emptyset \ne U \subset X$ open. For each $x \in U$ and $V \in \cn(x)$, $U \cap V \in \cn(x)$ by (F2). Thus there exists $y \in A \cap U \cap V$. By (3) of \ref{definition:closure}, $x \in \overline{A \cap U}$.
$(2) \Rightarrow (3)$: Let $\emptyset \ne U \subset X$ open. For each $x \in U$ and $V \in \cn(x)$, $U \cap V \in \cn(x)$ by (F2). Thus there exists $y \in A \cap U \cap V$. By (3) of \autoref{definition:closure}, $x \in \overline{A \cap U}$.
$(3) \Rightarrow (1)$: $X$ is open.
\end{proof}
@@ -92,7 +92,7 @@
Let $\seqf{X_j}$ be separable topological spaces, then $\prod_{j = 1}^n X_j$ is also separable.
\end{proposition}
\begin{proof}
Let $\seq{A_j}$ such that $A_j \subset X_j$ is countable and dense for each $1 \le j \le n$. By \ref{proposition:dense-product}, $\prod_{j = 1}^n A_j$ is countable and dense in $\prod_{j = 1}^n X_j$.
Let $\seq{A_j}$ such that $A_j \subset X_j$ is countable and dense for each $1 \le j \le n$. By \autoref{proposition:dense-product}, $\prod_{j = 1}^n A_j$ is countable and dense in $\prod_{j = 1}^n X_j$.
\end{proof}
@@ -101,7 +101,7 @@
Let $X$ be a topological space and $A \subset X$, then $(A^o)^c = \overline{A^c}$.
\end{lemma}
\begin{proof}
Let $x \in X$. By (3) of \ref{definition:interior}, $x \in (A^o)^c$ if and only if there exists no $U \in \cn(x)$ such that $U \subset A$. Thus $x \in (A^o)^c$ if and only if $U \cap A^c \ne \emptyset$ for all $U \in \cn(x)$. By (2) of \ref{definition:closure}, this is equivalent to $x \in \ol{A^c}$.
Let $x \in X$. By (3) of \autoref{definition:interior}, $x \in (A^o)^c$ if and only if there exists no $U \in \cn(x)$ such that $U \subset A$. Thus $x \in (A^o)^c$ if and only if $U \cap A^c \ne \emptyset$ for all $U \in \cn(x)$. By (2) of \autoref{definition:closure}, this is equivalent to $x \in \ol{A^c}$.
\end{proof}
@@ -120,9 +120,9 @@
\begin{proof}
$(1) \Rightarrow (2)$: $\cn(x) \supset \fB$.
$(2) \Rightarrow (3)$: By (2) of \ref{definition:closure}, $x \in \overline{A}$. Since $V \cap A^c \ne \emptyset$ for all $V \in \fB$, there exists no open set $U \subset A$ with $x \in A$. By (2) of \ref{definition:interior}, $x \not\in A^o$.
$(2) \Rightarrow (3)$: By (2) of \autoref{definition:closure}, $x \in \overline{A}$. Since $V \cap A^c \ne \emptyset$ for all $V \in \fB$, there exists no open set $U \subset A$ with $x \in A$. By (2) of \autoref{definition:interior}, $x \not\in A^o$.
$(3) \Rightarrow (4)$: By \ref{lemma:closurecomplement}.
$(3) \Rightarrow (4)$: By \autoref{lemma:closurecomplement}.
$(4) \Rightarrow (1)$: By (2) of \ref{definition:closure}.
$(4) \Rightarrow (1)$: By (2) of \autoref{definition:closure}.
\end{proof}

62
src/topology/main/lch.tex
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@@ -12,9 +12,9 @@
If the above holds, then $X$ is a \textbf{locally compact Hausdorff (LCH)} space.
\end{definition}
\begin{proof}
(1) $\Rightarrow$ (2): Let $K \in \cn(x)$ be compact and $U \in \cn(x)$, then $\overline{U \cap K}$ is closed. By \ref{proposition:compact-closed}, $K$ itself is closed, so $\overline{U \cap K} \subset K$ is a closed subset of a compact set, and compact by \ref{proposition:compact-extensions}.
(1) $\Rightarrow$ (2): Let $K \in \cn(x)$ be compact and $U \in \cn(x)$, then $\overline{U \cap K}$ is closed. By \autoref{proposition:compact-closed}, $K$ itself is closed, so $\overline{U \cap K} \subset K$ is a closed subset of a compact set, and compact by \autoref{proposition:compact-extensions}.
(2) $\Rightarrow$ (3): Let $U \in \cn(x)$, then there exists $K \in \cn(x)$ with $x \in K \subset U$. By \ref{proposition:compact-closed}, $K$ is closed, so $\overline{K^o} \subset K$ is compact by \ref{proposition:compact-extensions}.
(2) $\Rightarrow$ (3): Let $U \in \cn(x)$, then there exists $K \in \cn(x)$ with $x \in K \subset U$. By \autoref{proposition:compact-closed}, $K$ is closed, so $\overline{K^o} \subset K$ is compact by \autoref{proposition:compact-extensions}.
\end{proof}
\begin{lemma}
@@ -22,14 +22,16 @@
Let $X$ be a LCH space, $K \subset X$ be compact, and $U \in \cn(K)$, then there exits $V \in \cn^o(K)$ precompact such that $K \subset V \subset \ol{V} \subset U$.
\end{lemma}
\begin{proof}
For each $x \in K$, there exists $V_x \in \cn^o(x)$ be precompact such that $x \in V_x \subset \overline{V_x} \subset U$ by (3) of \ref{definition:lch}. Since $K$ is compact, there exists $\seqf{x_j} \subset K$ such that
For each $x \in K$, there exists $V_x \in \cn^o(x)$ be precompact such that $x \in V_x \subset \overline{V_x} \subset U$ by (3) of \autoref{definition:lch}. Since $K$ is compact, there exists $\seqf{x_j} \subset K$ such that
\[
K \subset \bigcup_{j = 1}^n V_{x_j} \subset U
\]
By \ref{proposition:closure-finite-union},
By \autoref{proposition:closure-finite-union},
\[
\ol{\bigcup_{j = 1}^n V_{x_j}} = \bigcup_{j = 1}^n \overline{V_{x_j}} \subset U
\]
so $V = \bigcup_{j = 1}^n V_{x_j} \in \cn^o(K)$ is precompact.
\end{proof}
@@ -38,20 +40,22 @@
Let $X$ be a LCH space, $K \subset X$ be compact, and $U \in \cn(K)$, then there exists $f \in C_c(X; [0, 1])$ such that $\supp{f} \subset U$.
\end{lemma}
\begin{proof}
By \ref{lemma:lch-compact-neighbour}, there exists $V, W \in \cn^o(K)$ precompact such that
By \autoref{lemma:lch-compact-neighbour}, there exists $V, W \in \cn^o(K)$ precompact such that
\[
K \subset V \subset \ol{V} \subset W \subset \ol{W} \subset U
\]
As $\ol{W}$ is compact, it is normal by \ref{proposition:compact-hausdorff-normal}. Since $X$ is Hausdorff, $K \subset \ol{W}$ is closed by \ref{proposition:compact-closed}.
By Urysohn's lemma (\ref{lemma:urysohn}), there exists $f \in C(\ol{V}; [0, 1])$ such that $f|_K = 1$ and $f|_{\ol{W} \setminus V} = 0$. Let
As $\ol{W}$ is compact, it is normal by \autoref{proposition:compact-hausdorff-normal}. Since $X$ is Hausdorff, $K \subset \ol{W}$ is closed by \autoref{proposition:compact-closed}.
By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $f \in C(\ol{V}; [0, 1])$ such that $f|_K = 1$ and $f|_{\ol{W} \setminus V} = 0$. Let
\[
F: X \to [0, 1] \quad x \mapsto \begin{cases}
f(x) &x \in W \\
0 &x \in X \setminus \ol{V}
\end{cases}
\]
then by the gluing lemma for continuous functions (\ref{lemma:gluing-continuous}), $F \in C_c(X; [0, 1])$ with $F|_{K} = 1$ and $\supp{f} \subset U$.
then by the \hyperref[gluing lemma for continuous functions]{lemma:gluing-continuous}, $F \in C_c(X; [0, 1])$ with $F|_{K} = 1$ and $\supp{f} \subset U$.
\end{proof}
\begin{theorem}[Tietze Extension Theorem (LCH)]
@@ -59,16 +63,17 @@
Let $X$ be a LCH space, $K \subset X$ be compact, $U \in \cn^o(K)$, and $f \in C(K; \real)$, then there exists $F \in C_c(U; \real)$ such that $F|_K = f$.
\end{theorem}
\begin{proof}
By \ref{lemma:lch-compact-neighbour}, there exists $V, W \in \cn^o(K)$ precompact such that $K \subset V \subset \ol{V} \subset U$. As $\ol{W}$ is compact, it is normal by \ref{proposition:compact-hausdorff-normal}. Since $X$ is Hausdorff, $K \subset \ol{W}$ is closed by \ref{proposition:compact-closed}.
By \autoref{lemma:lch-compact-neighbour}, there exists $V, W \in \cn^o(K)$ precompact such that $K \subset V \subset \ol{V} \subset U$. As $\ol{W}$ is compact, it is normal by \autoref{proposition:compact-hausdorff-normal}. Since $X$ is Hausdorff, $K \subset \ol{W}$ is closed by \autoref{proposition:compact-closed}.
By the Tietze extension theorem (\ref{theorem:tietze}), there exists $F \in C(\ol{W}; \real)$ such that $F|_K = f$. By Urysohn's lemma (\ref{lemma:lch-urysohn}), there exists $\eta \in C_c(X; [0, 1])$ such that $\eta|_K = 1$ and $\supp{\eta} \subset V$. In which case, define
By the \hyperref[Tietze extension theorem]{theorem:tietze}, there exists $F \in C(\ol{W}; \real)$ such that $F|_K = f$. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $\eta \in C_c(X; [0, 1])$ such that $\eta|_K = 1$ and $\supp{\eta} \subset V$. In which case, define
\[
\ol F: X \to \real \quad x \mapsto \begin{cases}
\eta(x) \cdot f(x) &x \in V \\
0 &x \in X \setminus \supp{\eta}
\end{cases}
\]
then by the gluing lemma for continuous functions (\ref{lemma:gluing-continuous}), $\ol F \in C_c(X; \real)$ with $\ol F|_K = F|_K = f$ and $\supp{F} \subset \supp{\eta} \subset V \subset U$.
then by the \hyperref[gluing lemma for continuous functions]{lemma:gluing-continuous}, $\ol F \in C_c(X; \real)$ with $\ol F|_K = F|_K = f$ and $\supp{F} \subset \supp{\eta} \subset V \subset U$.
\end{proof}
\begin{proposition}[{{\cite[Proposition 4.39]{Folland}}}]
@@ -88,10 +93,11 @@
\item[(b)] For each $0 \le k < n$, $\overline{U_k} \subset U_{k+1}$.
\item[(c)] For each $1 \le k \le n$, $U_k \supset \bigcup_{j = 1}^k K_j$.
\end{enumerate}
By \ref{lemma:lch-compact-neighbour}, there exists $U_{n+1} \in \cn^o(\overline{U_n} \cup K_{n+1})$ precompact. In which case, by (c),
By \autoref{lemma:lch-compact-neighbour}, there exists $U_{n+1} \in \cn^o(\overline{U_n} \cup K_{n+1})$ precompact. In which case, by (c),
\[
U_{n+1} \supset \ol{U_n} \cup K_{n+1} \supset \bigcup_{j = 1}^n K_j \cup K_{n+1} = \bigcup_{j = 1}^{n+1}K_j
\]
Thus $\bracs{U_j}_0^{n+1}$ satisfies (a), (b), and (c), and $\seq{U_n}$ is an exhaustion of $X$ by compact sets.
\end{proof}
@@ -108,14 +114,16 @@
\[
F_j = \bigcup_{\substack{1 \le k \le m \\ N_{x_k} \subset U_j}}N_{x_k}
\]
then $F_j \subset U_j$ is compact, and $\bigcup_{j = 1}^n F_j \supset K$.
By Urysohn's lemma (\ref{lemma:lch-urysohn}), there exists $\seqf{f_j} \subset C_c(X; [0, 1])$ such that for each $1 \le j \le n$, $f_j|_{F_j} = 1$, and $\supp{f_j} \subset U_j$.
By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $\seqf{f_j} \subset C_c(X; [0, 1])$ such that for each $1 \le j \le n$, $f_j|_{F_j} = 1$, and $\supp{f_j} \subset U_j$.
By Urysohn's lemma again, there exists $f_{j + 1} \in C(X; [0, 1])$ such that $f_{j+1}|_{K} = 0$ and $\bracs{f_{j+1} = 0} \subset \bigcup_{j = 1}^n \supp{f_j}$. Let $F = \sum_{j = 1}^{n+1}f_j$, then $F(x) > 0$ for all $x \in X$. For each $1 \le j \le n$, let $g_j = f_j/F$, then $g_j \in C_c(X; [0, 1])$ with $\supp{g_j} \subset U_j$. In addition, since $f_{j+1}|_K = 0$,
\[
\sum_{j = 1}^n g_j|_K = \frac{\sum_{j = 1}^n f_j}{F} = \frac{\sum_{j = 1}^n f_j}{\sum_{j = 1}^n f_j} = 1
\]
Therefore $\seqf{g_j}$ is the desired partition of unity.
\end{proof}
@@ -124,17 +132,19 @@
Let $X$ be a LCH space and $\ce \subset 2^X$ be a locally finite precompact open cover of $X$, then there exists locally finite precompact open covers $\bracs{F_E}_{E \in \ce}, \bracs{G_E}_{E \in \ce} \subset 2^X$ such that for each $E \in \ce$, $F_E \subset \ol{F_E} \subset E \subset \ol{E} \subset G_E$.
\end{lemma}
\begin{proof}
$(\bracs{F_E}_{E \in \ce})$: For each $E \in \ce$, $\bracs{F \in \ce|F \cap \ol E \ne \emptyset}$ is finite by \ref{lemma:locally-finite-compact}. Let
$(\bracs{F_E}_{E \in \ce})$: For each $E \in \ce$, $\bracs{F \in \ce|F \cap \ol E \ne \emptyset}$ is finite by \autoref{lemma:locally-finite-compact}. Let
\[
F_E = \bigcup_{\substack{F \in \ce} \\ F \cap \ol E \ne \emptyset}F
\]
then $F_E \in \cn(\ol{E})$ is precompact.
Let $N \subset X$ and $E \in \ce$. If $N \cap F_E \ne \emptyset$, then there exists $F \in \ce$ such that $N \cap F \ne \emptyset$ and $F \cap \ol{E} \ne \emptyset$. Thus
\[
\bracs{E \in \ce|N \cap F_E \ne \emptyset} \subset \bigcup_{\substack{F \in \ce \\ F \cap N \ne \emptyset}}\bracs{E \in \ce|F \cap \ol{E} \ne \emptyset} \subset \bigcup_{\substack{F \in \ce \\ F \cap N \ne \emptyset}}\bracs{E \in \ce|\ol{F} \cap \ol{E} \ne \emptyset}
\]
By \ref{lemma:locally-finite-closure}, $\bracsn{\ol E|E \in \ce}$ is also locally finite. Hence for every $F \in \ce$, $\bracsn{E \in \ce|F \cap \ol{E} \ne \emptyset}$ is finite.
By \autoref{lemma:locally-finite-closure}, $\bracsn{\ol E|E \in \ce}$ is also locally finite. Hence for every $F \in \ce$, $\bracsn{E \in \ce|F \cap \ol{E} \ne \emptyset}$ is finite.
Let $x \in X$, then there exists $N \in \cn(x)$ such that $\bracs{F \in \ce|N \cap F \ne \emptyset}$ is finite. In which case, $\bracs{E \in \ce|N \cap F_E \ne \emptyset}$ is finite as well. Therefore $\bracs{F_E}_{E \in \ce}$ is locally finite.
@@ -149,17 +159,20 @@
\[
G_E = \bigcup_{\substack{x \in X_\ce \\ \ol{N_x} \subset E}}N_x
\]
then $\bracs{G_E}_{E \in \ce}$ is an open cover of $X$. Since $G_E \subset E$ for all $E \in \ce$, $\bracs{G_E}_{E \in \ce}$ is locally finite.
It remains to show that $\ol{G_E} \subset E$. Let $x \in X_F$ such that $N_x \subset E$, then $N_x \cap F \ne \emptyset$. Since $N_x \subset E$, $E \cap F \ne \emptyset$. Thus
\[
\bracsn{x \in X_\ce|\ol{N_x} \subset E} \subset \bigcup_{\substack{F \in \ce \\ E \cap F \ne \emptyset}}X_F \subset \bigcup_{\substack{F \in \ce \\ \ol E \cap F \ne \emptyset}}X_F
\]
is finite by \ref{lemma:locally-finite-compact}, so
is finite by \autoref{lemma:locally-finite-compact}, so
\[
\ol{G_E} = \bigcup_{\substack{x \in X_\ce \\ \ol{N_x} \subset E}}\ol N_x \subset E
\]
by \ref{proposition:closure-finite-union}.
by \autoref{proposition:closure-finite-union}.
\end{proof}
\begin{proposition}
@@ -177,31 +190,34 @@
\begin{proof}
(1) $\Rightarrow$ (2): For each $x \in X$, there exists a precompact open neighbourhood $U_x \in \cn^o(x)$. Since $\bracs{U_x| x \in X}$ is an open cover of $X$, there exists a locally finite refinement $\mathcal{V}$. For each $V \in \mathcal{V}$, there exists $x \in X$ such that $V \subset U_x$. In which case, $\ol{V} \subset \ol{U_x}$ is compact.
(2) $\Rightarrow$ (3): Let $\cf \subset 2^X$ be a locally finite open cover of $X$ consisting of precompact open sets. By \ref{lemma:lch-locally-finite-precompact-refine}, there exists a locally finite open cover $\bracs{G_F}_{F \in \cf}$ of $X$ consisting of precompact open sets such that $\ol{F} \subset G_F$ for all $F \in \cf$.
(2) $\Rightarrow$ (3): Let $\cf \subset 2^X$ be a locally finite open cover of $X$ consisting of precompact open sets. By \autoref{lemma:lch-locally-finite-precompact-refine}, there exists a locally finite open cover $\bracs{G_F}_{F \in \cf}$ of $X$ consisting of precompact open sets such that $\ol{F} \subset G_F$ for all $F \in \cf$.
For each $F \in \cf$, let
\[
\mathcal{U}_F = \bracs{U \cap G_F|U \in \mathcal{U}}
\]
then $\mathcal{U}_F$ is a precompact open cover of $\ol{F}$. By compactness of $\ol{F}$, there exists $\mathcal{V}_F \subset \mathcal{U}_F$ finite such that $\ol{F} \subset \bigcup_{V \in \mathcal{V}_F}V$.
Let $\mathcal{V} = \bigcup_{F \in \cf}\mathcal{V}_F$, then $\mathcal{V}$ is a precompact open cover of $X$. For any $x \in X$, there exists $N \in \cn(x)$ such that $\bracs{F \in \cf|N \cap G_F}$ is finite. Thus
\[
\bracs{V \in \mathcal{V}| N \cap V} \subset \bigcup_{\substack{F \in \cf \\ N \cap G_F \ne \emptyset}}\mathcal{V}_F
\]
is finite, and $\mathcal{V}$ is locally finite.
(3) $\Rightarrow$ (4): By \ref{lemma:lch-locally-finite-precompact-refine}.
(3) $\Rightarrow$ (4): By \autoref{lemma:lch-locally-finite-precompact-refine}.
(4) $\Rightarrow$ (5): Let $\seqi{V}, \seqi{W} \subset 2^X$ be locally finite refinements of $\mathcal{U}$ consisting of precompact open sets such that for each $i \in I$, $\ol{W_i} \subset V_i$.
By Urysohn's Lemma (\ref{lemma:lch-urysohn}), there exists $\seqi{f} \in C_c(X; [0, 1])$ such that for each $i \in I$, $f_i|_{\ol{W_i}} = 1$ and $\supp{f_i} \subset V_i$.
By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $\seqi{f} \in C_c(X; [0, 1])$ such that for each $i \in I$, $f_i|_{\ol{W_i}} = 1$ and $\supp{f_i} \subset V_i$.
Let $F = \sum_{i \in I}f_i$. For each $x \in X$, there exists $N_x \in \cn^o(x)$ such that $\bracs{i \in I|N_x \cap V_i \ne \emptyset}$ is finite. In which case,
\[
F|_{N_x} = \sum_{\substack{i \in I \\ N_x \cap V_i \ne \emptyset}}f_i|_{N_x}
\]
thus $F|_{N_x} \in C(N_x; \real)$. By \ref{lemma:gluing-continuous}, $F \in C(X; \real)$.
thus $F|_{N_x} \in C(N_x; \real)$. By \autoref{lemma:gluing-continuous}, $F \in C(X; \real)$.
Since $\seqi{W}$ is an open cover of $X$, $F(x) > 0$ for all $x \in X$. For each $i \in I$, let $g_i = f_i/F$, then $g_i \in C_c(X; [0, 1])$ with $\supp{g_i} = \supp{f_i} \subset W_i$. For any $x \in X$, there exists $N_x \in \cn^o(x)$ such that $\bracs{i \in I|N_x \cap W_i \ne \emptyset}$ is finite. In which case, $\bracs{i \in I|0 < g_i|_{N_x}}$ is also finite. Thus $\seqi{g}$ is a $C_c$ partition of unity subordinate to $\mathcal{U}$.
@@ -217,7 +233,7 @@
Let $X$ be a $\sigma$-compact LCH space, then $X$ is paracompact.
\end{proposition}
\begin{proof}
By \ref{proposition:lch-sigma-compact}, there exists an exhaustion $\seq{U_n} \subset 2^X$ of $X$ by precompact open sets. Denote $U_0 = \emptyset$. For each $n \in \natp$, let $V_n = U_{n+1} \setminus \ol{U_{n-1}}$.
By \autoref{proposition:lch-sigma-compact}, there exists an exhaustion $\seq{U_n} \subset 2^X$ of $X$ by precompact open sets. Denote $U_0 = \emptyset$. For each $n \in \natp$, let $V_n = U_{n+1} \setminus \ol{U_{n-1}}$.
Let $x \in X$, then there exists $n \in \natp$ such that $x \in U_n \setminus U_{n-1}$. In which case, if $n > 1$, then $x \in U_{n} \setminus \ol{U_{n - 2}} = V_{n-1}$. If $n = 1$, then $x \in U_{2} = V_1$. Thus $\seq{V_n}$ is an open cover of $X$. In addition, for any $m, n \in \natp$ with $m \le n$, $V_m \cap V_n \ne \emptyset$ implies that $n - m < 2$, so $\seq{V_n}$ is locally finite. By (2) of \ref{proposition:lch-paracompact}, $X$ is paracompact.
Let $x \in X$, then there exists $n \in \natp$ such that $x \in U_n \setminus U_{n-1}$. In which case, if $n > 1$, then $x \in U_{n} \setminus \ol{U_{n - 2}} = V_{n-1}$. If $n = 1$, then $x \in U_{2} = V_1$. Thus $\seq{V_n}$ is an open cover of $X$. In addition, for any $m, n \in \natp$ with $m \le n$, $V_m \cap V_n \ne \emptyset$ implies that $n - m < 2$, so $\seq{V_n}$ is locally finite. By (2) of \autoref{proposition:lch-paracompact}, $X$ is paracompact.
\end{proof}

2
src/topology/main/local-path-connected.tex
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@@ -17,7 +17,7 @@
\end{enumerate}
\end{proposition}
\begin{proof}
(1): Let $A \subset X$ be a path component and $x \in A$, then there exists $U \in \cn(x)$ connected. By \ref{proposition:path-connected-union}, $A \cup U \in \cn(x)$ is also connected. Since $A$ is a path-component, $A \cup U \subset A \in \cn(x)$. Thus $A$ is open by \ref{lemma:openneighbourhood}.
(1): Let $A \subset X$ be a path component and $x \in A$, then there exists $U \in \cn(x)$ connected. By \autoref{proposition:path-connected-union}, $A \cup U \in \cn(x)$ is also connected. Since $A$ is a path-component, $A \cup U \subset A \in \cn(x)$. Thus $A$ is open by \autoref{lemma:openneighbourhood}.
(2): Let $P$ be a path component in $X$ and $C \supset P$ be its connected components. If $C$ is not path-connected, then $P \subsetneq C$ there exists path-components $\seqi{P} \subset 2^C$ such that $C = \bigsqcup_{i \in I}P_i$. In which case, $C \setminus P$ is open by (1), and $C$ is not connected, which is impossible.

5
src/topology/main/neighbourhoods.tex
View File

@@ -50,6 +50,7 @@
\[
\topo = \bracs{U \subset X| U \in \cn(x) \forall x \in U}
\]
Firstly, $\emptyset$ satisfies the condition vacuously, so $\emptyset \in \topo$.
For any $x \in X$, $\cn(x)$ is non-empty and there exists $V \in \cn(x)$. Since $X \supset V$, $X \in \topo$ by (F1).
@@ -63,11 +64,13 @@
\[
U = \bracs{y \in V: V \in \cn(y)}
\]
then $U \supset U_0$ and $U \in \cn(x)$ by (V1). Let $y \in U$, then (V4) implies that there exists $W \in \cn(y)$ such that $V \in \cn(z)$ for all $z \in W$. Thus $W \subset U$ and $U \in \cn(y)$ by (F1).
\textit{Uniqueness}: Let $\mathcal{R}$ be a topology such that $\cn_{\mathcal{R}} = \cn$, then by \ref{lemma:openneighbourhood},
\textit{Uniqueness}: Let $\mathcal{R}$ be a topology such that $\cn_{\mathcal{R}} = \cn$, then by \autoref{lemma:openneighbourhood},
\[
\mathcal{R} = \bracs{U \subset X| U \in \cn_{\mathcal{R}}(x) \forall x \in U} = \bracs{U \subset X| U \in \cn(x) \forall x \in U} = \topo
\]
so $\topo$ is unique.
\end{proof}

13
src/topology/main/normal.tex
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@@ -30,16 +30,17 @@
\begin{proof}
(1): Let $\seq{q_n}$ be an enumeration of $\rational \cap (0, 1)$. For each $n \in \natp$, let $Q_n = \bracs{0, 1} \cup \bracs{q_k|1 \le k \le n}$.
Let $U_1 = B^c$. By (2) of \ref{definition:topology-normal}, there exists $U_0 \in \cn(A)$ such that $A \subset U_0 \subset \ol{U_0} \subset B^c$. In which case, for $n = 0$,
Let $U_1 = B^c$. By (2) of \autoref{definition:topology-normal}, there exists $U_0 \in \cn(A)$ such that $A \subset U_0 \subset \ol{U_0} \subset B^c$. In which case, for $n = 0$,
\begin{enumerate}
\item[(i)] $U_1 = B^c$.
\item[(ii)] For any $p, q \in Q_k$ with $p < q$, $\overline{U_p} \subset U_q$.
\end{enumerate}
Suppose inductively that $\bracs{U_q|q \in Q_n}$ has been constructed, and (ii) holds for $n$. Let $p = \max\bracs{r \in Q_n|r < q_{n+1}}$ and $q = \min\bracs{r \in Q_n|r > r_{n+1}}$. By (2) of \ref{definition:topology-normal}, there exists $U_{q_{n+1}} \in \cn^o(\overline{U_p})$ such that
Suppose inductively that $\bracs{U_q|q \in Q_n}$ has been constructed, and (ii) holds for $n$. Let $p = \max\bracs{r \in Q_n|r < q_{n+1}}$ and $q = \min\bracs{r \in Q_n|r > r_{n+1}}$. By (2) of \autoref{definition:topology-normal}, there exists $U_{q_{n+1}} \in \cn^o(\overline{U_p})$ such that
\[
U_p \subset \overline{U_p} \subset U_{q_{n+1}} \subset \overline{U_{q_{n+1}}} \subset U_q
\]
and $\bracs{U_q|q \in Q_{n+1}}$ satisfies (ii) for $n + 1$.
Now suppose that $\bracs{U_q|q \in \rational \cap [0, 1]}$ has been constructed and (ii) holds for all $n \in \nat$. For any $p, q \in \rational \cap [0, 1]$ with $p < q$, there exists $n \in \nat$ such that $p, q \in Q_n$. In which case, $\ol{U_p} \subset U_q$ by (ii). Thus (b) holds.
@@ -48,16 +49,19 @@
\[
f: X \to [0, 1] \quad x \mapsto \inf\bracs{q \in [0, 1] \cap \rational| x \in U_q}
\]
where $f(x) = 1$ if $x \not\in \bigcup_{q \in [0, 1] \cap \rational}U_q$. Since $A \subset \bigcap_{q \in [0, 1] \cap \rational}U_q$ and $U_1 = B^c$, $f|_A = 0$ and $f|_B = 1$.
Let $\alpha \in [0, 1]$, then
\[
f^{-1}([0, \alpha)) = \bigcup_{\substack{q \in \rational \cap [0, 1] \\ q < \alpha}}U_q
\]
is open. On the other hand, let $x \in X$, then $f(x) > \alpha$ if and only if there exists $q \in (\alpha, f(x)) \cap \rational$ such that $x \not\in U_q$. By (b) of (1), this is equivalent to the existence of $p \in (\alpha, f(x)) \cap \rational$ such that $x \not\in \ol{U_p}$. In which case,
\[
f^{-1}((b, 1]) = \bigcup_{\substack{q \in \rational \cap [0, 1] \\ q > \alpha}}\overline{U_p}^c
\]
is open.
\end{proof}
@@ -70,18 +74,21 @@
\[
R: BC(X; \real) \to BC(A; \real) \quad g \mapsto g|_A
\]
then $R \in L(BC(X; \real); BC(A; \real))$.
For any $g \in C(A; [0, 1])$, let
\[
B = g^{-1}(\norm{g}_u \cdot [0, 1/3]) \quad C = g^{-1}(\norm{g}_u \cdot [2/3, 1])
\]
then $B, C \subset A$ are closed with $B \cap C = \emptyset$. Since $A$ is closed, $B, C \subset X$ are closed. By Urysohn's lemma, there exists $h \in C(X; [0, 1/3])$ such that $h|_C = 1/3$ and $h|_B = 0$. Thus $g - h|_A \in C(A; [0, 2/3])$.
By linearity, this implies that for any $g \in BC(A; \real)$, there exists $h \in BC(A; \real)$ such that $\norm{h}_u \le \norm{g}_u/3$ and $\norm{g - h|_A}_u \le 2\norm{g}_u/3$.
Since $\real$ is complete, so is $BC(X; \real)$ by \ref{proposition:set-uniform-complete}. Using successive approximations (\ref{theorem:successive-approximation}), for every $f \in BC(A; \real)$, there exists $F \in BC(X; \real)$ such that $RF = F|_A = f$ and
Since $\real$ is complete, so is $BC(X; \real)$ by \autoref{proposition:set-uniform-complete}. Using \hyperref[successive approximations]{theorem:successive-approximation}, for every $f \in BC(A; \real)$, there exists $F \in BC(X; \real)$ such that $RF = F|_A = f$ and
\[
\norm{F}_u \le \frac{1}{3} \cdot \frac{1}{1 - 2/3} \cdot \norm{f}_u = \norm{f}_u
\]
\end{proof}

1
src/topology/main/para.tex
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@@ -15,6 +15,7 @@
\[
\bracs{U \in \mathcal{U}|U \cap K \ne \emptyset} \subset \bigcup_{x \in X_K}\bracs{U \in \mathcal{U}|U \cap N_x \ne \emptyset}
\]
\end{proof}
\begin{lemma}

6
src/topology/main/path-connected.tex
View File

@@ -18,10 +18,11 @@
\begin{proof}
Let $U, V \subset [0, 1]$ be open with $[0, 1] = U \cup V$. If $\sup U = \sup V$, then $\sup U = \sup V = 1$ and $U, V \in \cn^o(1)$, so $U \cap V \ne \emptyset$. If $\sup U < \sup V \le 1$, then $x \not\in U$ and $x \in V$. In which case, $V \in \cn^o(x)$ and $V \cap U \ne \emptyset$. Therefore $[0, 1]$ is connected.
Fix $x \in X$. For any $y \in X$, let $f_y \in C([0, 1]; X)$ be a path from $x$ to $y$, then $f_y([0, 1])$ is connected with $x, y \in f_y([0, 1])$ by \ref{proposition:connected-image}. By \ref{proposition:connected-union},
Fix $x \in X$. For any $y \in X$, let $f_y \in C([0, 1]; X)$ be a path from $x$ to $y$, then $f_y([0, 1])$ is connected with $x, y \in f_y([0, 1])$ by \autoref{proposition:connected-image}. By \autoref{proposition:connected-union},
\[
X = \bigcup_{y \in X}f_y([0, 1])
\]
is connected.
\end{proof}
@@ -37,6 +38,7 @@
g(2(t - 1/2)) &t \in [1/2, 1]
\end{cases}
\]
is a path from $y$ to $z$.
\end{proof}
@@ -45,5 +47,5 @@
Let $X$ be a topological space and $A \subset X$ be path-connected, then there exists a unique path-connected set $C \supset A$ such that for any $C' \supset A$ path-connected, $C \supset C'$. The set $C$ is the \textbf{path-component} of $A$.
\end{definition}
\begin{proof}
Let $C$ be the union of all path-connected sets containing $A$, then $C$ is path-connected by \ref{proposition:path-connected-union} and the maximum path-connected set containing $A$ by definition.
Let $C$ be the union of all path-connected sets containing $A$, then $C$ is path-connected by \autoref{proposition:path-connected-union} and the maximum path-connected set containing $A$ by definition.
\end{proof}

18
src/topology/main/product.tex
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@@ -11,6 +11,7 @@
\cb(\ce) = \bracs{\bigcap_{k = 1}^n \pi_{i_k}^{-1}(U_{k}) \bigg | U_{k} \in \topo_{i_k}, \seqf{i_k} \subset I, n \in \nat^+}
\]
is a base for $\topo$.
\item[(U)] For any topological space space $Y$ and $\seqi{f}$ where $f_i \in C(Y; X_i)$ for all $i \in I$, there exists a unique $f \in C(Y; X)$ such that the following diagram commutes
@@ -22,16 +23,18 @@
}
\]
for all $i \in I$.
\end{enumerate}
\end{definition}
\begin{proof}
(1): By (3) of \ref{definition:initial-topology}.
(1): By (3) of \autoref{definition:initial-topology}.
(U): Let $f \in \prod_{i \in I}f_i$, then $f$ is the unique function such that the diagrams commute. For each $\bigcap_{k = 1}^n \pi_{i_k}^{-1}(U_k) \in \topo$,
\[
f^{-1}\paren{\bigcap_{k = 1}^n \pi_{i_k}^{-1}(U_k)} = \bigcap_{k = 1}^n f_{i_k}^{-1}(U_k)
\]
which is open in $Y$ by (O2).
\end{proof}
@@ -46,6 +49,7 @@ Let $\bracsn{(X_i, \topo_i)}_{i \in I}$ be a family of topological spaces and $\
\[
B \subset \bigcap_{k = 1}^n B_k \subset \bigcap_{k = 1}^n \pi_{i_k}^{-1}(U_k) \subset U
\]
Therefore $\fB$ converges to $x$.
\end{proof}
@@ -62,23 +66,26 @@ Let $\bracsn{(X_i, \topo_i)}_{i \in I}$ be a family of topological spaces and $\
}
\]
for all $i \in I$.
\item If $X$ is T0 and equipped with the initial topology induced by $\seqi{f}$, then $f$ is an embedding.
\end{enumerate}
\end{proposition}
\begin{proof}
(1): By (U) of \ref{definition:product-topology}.
(1): By (U) of \autoref{definition:product-topology}.
(2): Let $x, y \in X$ with $x \ne y$. Assume without loss of generality that there exists $U \in \cn(x)$ such that $y \not\in U$. Let $J \subset I$ finite, $\seqj{U}$, and $\seqj{V}$ such that
\[
U = \bigcap_{j \in J}f_j^{-1}(U_j) = f^{-1}\paren{\bigcap_{j \in J}\pi_j^{-1}(U_j)}
\]
then $f(x) \in \bigcap_{j \in J}\pi_j^{-1}(U_j)$ but $f(y) \not\in \bigcap_{j \in J}\pi_j^{-1}(U_j)$. Thus $f$ is injective.
Let $U \subset X$ be open. Assume without loss of generality that there exists $J \subset I$ finite and $\bracs{U_j}_{j \in J}$ such that $U = \bigcap_{j \in J}f_j^{-1}(U_j)$. In which case, $\bigcap_{j \in J}\pi_j^{-1}(U_j)$ is open in $\prod_{i \in I}Y_i$ and
\[
f(U) = f(X) \cap \bigcap_{j \in J}\pi_j^{-1}(U_j)
\]
is a relatively open set.
\end{proof}
@@ -95,13 +102,14 @@ Let $\bracsn{(X_i, \topo_i)}_{i \in I}$ be a family of topological spaces and $\
}
\]
for all $i \in I$.
\item $f$ is an embedding.
\end{enumerate}
\end{proposition}
\begin{proof}
(1): By (U) of \ref{definition:product-topology}/\ref{definition:product-uniform}.
(1): By (U) of \autoref{definition:product-topology}/\autoref{definition:product-uniform}.
(2): Consider the following diagram
\[
@@ -112,11 +120,13 @@ Let $\bracsn{(X_i, \topo_i)}_{i \in I}$ be a family of topological spaces and $\
\prod_{i \in I} X_i \ar@{->}[r]_{\pi_i} \ar@{->}[u]^{f} & X_i \ar@{^{(}->}[u]_{f_i}
}
\]
Since each $X_i \to Y_i$ is an embedding, the composition
\[
\xymatrix{
\iota_P(\prod_{i \in I}X_i) \ar@{->}[r]^{\pi_i} & Y_i \ar@{->}[r]^{f_i^{-1}} & X_i
}
\]
is continuous/uniformly continuous. By (U) of \ref{definition:product-topology}/\ref{definition:product-uniform}, $f$ is an embedding.
is continuous/uniformly continuous. By (U) of \autoref{definition:product-topology}/\autoref{definition:product-uniform}, $f$ is an embedding.
\end{proof}

1
src/topology/main/quotient.tex
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@@ -35,6 +35,7 @@
\td X \ar@{->}[r]_{\tilde f} & Y
}
\]
\item $\pi$ is a quotient map.
\end{enumerate}
The space $(\td X, \pi)$ is the \textbf{quotient} of $X$ by $\sim$.

5
src/topology/main/regular.tex
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@@ -27,11 +27,12 @@
\begin{proof}
$(\Rightarrow)$: Let $x \in X$ and $\fF = \bracs{U \cap A| U \in \cn(x)}$, then $f(\fF) = F(\fF)$. By continuity of $F$, $F(\fF)$ converges to $F(x)$, so $\lim_{y \to x, x \in A}f(x)$ exists.
$(\Leftarrow)$: Let $F(x) = \lim_{y \to x, y \in A}f(y)$, then $F$ is well-defined and $F|_A = f$ by (4) of \ref{definition:hausdorff}.
$(\Leftarrow)$: Let $F(x) = \lim_{y \to x, y \in A}f(y)$, then $F$ is well-defined and $F|_A = f$ by (4) of \autoref{definition:hausdorff}.
Let $x \in X$ and $V \in \cn(F(x))$. Using (2) of \ref{definition:regular}, assume without loss of generality that $V$ is closed. By assumption, there exists $U \in \cn^o(x)$ such that $f(U \cap A) \subset V$. In which case, \ref{lemma:openneighbourhood} implies that $U \in \cn^o(y)$ for all $y \in U$. Since every limit point of a filter is a cluster point,
Let $x \in X$ and $V \in \cn(F(x))$. Using (2) of \autoref{definition:regular}, assume without loss of generality that $V$ is closed. By assumption, there exists $U \in \cn^o(x)$ such that $f(U \cap A) \subset V$. In which case, \autoref{lemma:openneighbourhood} implies that $U \in \cn^o(y)$ for all $y \in U$. Since every limit point of a filter is a cluster point,
\[
F(y) = \lim_{\substack{z \to y \\ z \in A}}f(z) \in \ol{f(U \cap A)} \subset V
\]
as $V$ is closed. Therefore $F(U) \subset V$, and $F$ is continuous.
\end{proof}