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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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35
src/topology/uniform/definition.tex
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@@ -8,6 +8,7 @@
\[
U^{-1} = \bracs{(y, x)| (x, y) \in U}
\]
A set $U \subset X \times X$ is \textbf{symmetric} if $U = U^{-1}$.
\end{definition}
@@ -17,6 +18,7 @@
\[
U \circ V = \bracs{(x, z) \in X \times X| \exists y \in Y: (x, y) \in U, (y, z) \in V}
\]
\end{definition}
\begin{definition}[Slice]
@@ -25,6 +27,7 @@
\[
U(A) = \bracs{y \in Y: (x, y) \in U, x \in A}
\]
is the \textbf{slice} of $U$ at $A$.
\end{definition}
@@ -50,6 +53,7 @@
\[
\fU_A = \bracs{U \cap (A \times A)| U \in \fU}
\]
forms a uniformity on $A$, known as the \textbf{subspace uniformity} induced on $A$.
\end{definition}
@@ -80,6 +84,7 @@
\[
\fU = \bracs{U \subset X \times X| \exists V \in \fB: V \subset U}
\]
\end{proposition}
\begin{proof}
(F1): By definition of $\fU$.
@@ -96,6 +101,7 @@
\[
\fB_S = \bracsn{U \cap U^{-1}| U \in \fB}
\]
is also a fundamental system of entourages.
\end{lemma}
\begin{proof}
@@ -103,6 +109,7 @@
\[
U \supset V \supset V \cap V^{-1} \in \fB_S
\]
so $\fb_S$ is a fundamental system of entourages.
\end{proof}
@@ -112,20 +119,23 @@
\[
\cn: X \to 2^X \quad x \mapsto \bracsn{U(x)| U \in \fU}
\]
then there exists a unique topology $\topo \subset 2^X$ such that $\cn_\topo = \cn$, known as the \textbf{topology induced by the uniform structure $\fU$}.
\end{definition}
\begin{proof}
Using \ref{proposition:neighbourhoodcharacteristic}, it is sufficient to show that $\cn(x)$ is non-empty for all $x \in X$, and that it satisfies (F1), (F2), (V1), and (V2). Firstly, since $\fU \ne \emptyset$, $\cn(x) \ne \emptyset$ for all $x \in X$.
Using \autoref{proposition:neighbourhoodcharacteristic}, it is sufficient to show that $\cn(x)$ is non-empty for all $x \in X$, and that it satisfies (F1), (F2), (V1), and (V2). Firstly, since $\fU \ne \emptyset$, $\cn(x) \ne \emptyset$ for all $x \in X$.
(F1): Let $U \in \fU$ and $V \supset U(x)$, then
\[
W = U \cup (\bracs{x} \times V) \supset U
\]
As $\fU$ satisfies (F1), $W \in \fU$. Thus
\[
V = (\bracs{x} \times V)(x) = (U \cup (\bracs{x} \times V))(x) = W(x) \in \cn(x)
\]
(F2): Let $U, V \in \fU$, then $U(x) \cap V(x) = (U \cap V)(x)$. As $\fU$ satisfies (F2), $U \cap V \in \fU$ and $(U \cap V)(x) \in \cn(x)$.
(V1): Let $U \in \fU$. By (U1), $\Delta \subset U$, so $x \in U(x)$.
@@ -170,9 +180,9 @@ V = (\bracs{x} \times V)(x) = (U \cup (\bracs{x} \times V))(x) = W(x) \in \cn(x)
with respect to the product topology on $X \times X$.
\end{proposition}
\begin{proof}
(1): Let $(x, y) \in V \circ M \circ V$. By \ref{lemma:compositiongymnastics}, there exists $(p, q) \in M$ such that $(x, y) \in V(p) \times V(q) \in \cn(p, q)$. In particular, if $(x, y) \in M$, then this implies that $V \circ M \circ V \in \cn(x, y)$. Thus $V \circ M \circ V \in \cn(M)$.
(1): Let $(x, y) \in V \circ M \circ V$. By \autoref{lemma:compositiongymnastics}, there exists $(p, q) \in M$ such that $(x, y) \in V(p) \times V(q) \in \cn(p, q)$. In particular, if $(x, y) \in M$, then this implies that $V \circ M \circ V \in \cn(x, y)$. Thus $V \circ M \circ V \in \cn(M)$.
(2): Let $(x, y) \in X \times X$. By \ref{lemma:compositiongymnastics}, the following are equivalent:
(2): Let $(x, y) \in X \times X$. By \autoref{lemma:compositiongymnastics}, the following are equivalent:
\begin{enumerate}
\item[(a)] $(x, y) \in V \circ M \circ V$ for all $V \in \fB$.
\item[(b)] For every $V \in \fB$, there exists $(p, q) \in M$ such that $(p, q) \in V(x) \times V(y)$.
@@ -183,6 +193,7 @@ V = (\bracs{x} \times V)(x) = (U \cup (\bracs{x} \times V))(x) = W(x) \in \cn(x)
\[
\ol{M} = \bigcap_{V \in \fB}V \circ M \circ V
\]
\end{proof}
\begin{proposition}[{{\cite[Corollary 2.1.1]{Bourbaki}}}]
@@ -203,7 +214,7 @@ V = (\bracs{x} \times V)(x) = (U \cup (\bracs{x} \times V))(x) = W(x) \in \cn(x)
\item[(c)] For all $V \in \fB$, $V(x) \cap A \ne \emptyset$.
\end{enumerate}
Since $\bracs{V(y): V \in \fB}$ is a fundamental system of neighbourhoods at $y$ (\ref{lemma:symmetricfundamentalentourage}), (c) is equivalent to $x \in \overline{A}$. Therefore $\ol{A} = \bigcap_{U \in \fB}U(A)$.
Since $\bracs{V(y): V \in \fB}$ is a fundamental system of neighbourhoods at $y$ (\autoref{lemma:symmetricfundamentalentourage}), (c) is equivalent to $x \in \overline{A}$. Therefore $\ol{A} = \bigcap_{U \in \fB}U(A)$.
\end{proof}
\begin{proposition}[{{\cite[Corollary 2.1.2]{Bourbaki}}}]
@@ -213,19 +224,21 @@ V = (\bracs{x} \times V)(x) = (U \cup (\bracs{x} \times V))(x) = W(x) \in \cn(x)
\item $\mathfrak{O} = \bracs{U^o| U \in \fU}$
\item $\mathfrak{K} = \bracsn{\overline{U}| U \in \fU}$.
\end{enumerate}
By \ref{lemma:symmetricfundamentalentourage}, there exists fundamental systems of entourages for $\fU$ consisting of symmetric and open/closed sets.
By \autoref{lemma:symmetricfundamentalentourage}, there exists fundamental systems of entourages for $\fU$ consisting of symmetric and open/closed sets.
\end{proposition}
\begin{proof}
Let $U \in \fU$, then there exists a symmetric entourage $V \in \fU$ such that $V \circ V \circ V \subset U$ by (U2) and \ref{lemma:symmetricfundamentalentourage}. By (1) of \ref{proposition:uniformneighbourhood}, $V \circ V \circ V \in \cn(V)$. Since
Let $U \in \fU$, then there exists a symmetric entourage $V \in \fU$ such that $V \circ V \circ V \subset U$ by (U2) and \autoref{lemma:symmetricfundamentalentourage}. By (1) of \autoref{proposition:uniformneighbourhood}, $V \circ V \circ V \in \cn(V)$. Since
\[
V \subset (V \circ V \circ V)^o \subset V \circ V \circ V \subset U
\]
the interior $(V \circ V \circ V)^o \in \fU$, and $U$ contains the interior of an entourage. Thus (1) is a fundamental system of entourages.
On the other hand, by (2) of \ref{proposition:uniformneighbourhood},
On the other hand, by (2) of \autoref{proposition:uniformneighbourhood},
\[
V \subset \overline{V} \subset V \circ V \circ V \subset U
\]
So $\overline{V} \in \fU$ and is contained in $U$. Therefore (2) is also a fundamental system of entourages.
\end{proof}
@@ -242,7 +255,7 @@ V = (\bracs{x} \times V)(x) = (U \cup (\bracs{x} \times V))(x) = W(x) \in \cn(x)
Let $X$ be a uniform space and $x \in X$, then the closed neighbourhoods of $x$ form a fundamental system of neighbourhoods at $x$.
\end{proposition}
\begin{proof}
By \ref{proposition:goodentourages} and \ref{lemma:openentourageneighbourhoods}, the closed neighbourhoods form a fundamental system of neighbourhoods.
By \autoref{proposition:goodentourages} and \autoref{lemma:openentourageneighbourhoods}, the closed neighbourhoods form a fundamental system of neighbourhoods.
\end{proof}
\begin{definition}[Separated]
@@ -260,9 +273,9 @@ V = (\bracs{x} \times V)(x) = (U \cup (\bracs{x} \times V))(x) = W(x) \in \cn(x)
\begin{proof}
$(1) \Rightarrow (5)$: Let $x, y \in X$ with $x \ne y$. Assume without loss of generality that there exists $U(x) \in \cn(x)$ such that $y \not\in U$. In which case, $(x, y) \not\in U$ and $\Delta \supset \bigcap_{U \in \fU}U$.
$(5) \Rightarrow (2)$: By \ref{proposition:goodentourages}, $\ol \Delta \subset \bigcap_{U \in \fU}\ol U = \Delta$, so $\ol \Delta$ is closed. By (6) of \ref{definition:hausdorff}, $X$ is Hausdorff.
$(5) \Rightarrow (2)$: By \autoref{proposition:goodentourages}, $\ol \Delta \subset \bigcap_{U \in \fU}\ol U = \Delta$, so $\ol \Delta$ is closed. By (6) of \autoref{definition:hausdorff}, $X$ is Hausdorff.
$(1) \Rightarrow (4)$: $X$ is T1 and satisfies (2) of \ref{definition:regular} by \ref{proposition:uniform-neighbourhoods}, so $X$ is regular.
$(1) \Rightarrow (4)$: $X$ is T1 and satisfies (2) of \autoref{definition:regular} by \autoref{proposition:uniform-neighbourhoods}, so $X$ is regular.
$(4) \Rightarrow (3) \Rightarrow (2) \Rightarrow (1)$: (T3) $\Rightarrow $ (T2) $\Rightarrow$ (T1) $\Rightarrow$ (T0).
\end{proof}
@@ -273,5 +286,5 @@ V = (\bracs{x} \times V)(x) = (U \cup (\bracs{x} \times V))(x) = W(x) \in \cn(x)
Let $(X, \fU)$ be a uniform space and $A \subset X$ be a dense subset, then $\bracsn{\overline{U}: U \in \fU_A}$ forms a fundamental system of entourages for $X$.
\end{proposition}
\begin{proof}
Let $U \in \fU$ be an open entourage, then by (3) of \ref{definition:dense}, $\overline{U \cap (A \times A)} = \overline{U}$ for all $U \in \fU$, so $\overline{U \cap (A \times A)}$ is an entourage. By \ref{proposition:goodentourages}, every closed entourage of $X$ contains an element of $\bracsn{\overline{U}: U \in \fU_A}$.
Let $U \in \fU$ be an open entourage, then by (3) of \autoref{definition:dense}, $\overline{U \cap (A \times A)} = \overline{U}$ for all $U \in \fU$, so $\overline{U \cap (A \times A)}$ is an entourage. By \autoref{proposition:goodentourages}, every closed entourage of $X$ contains an element of $\bracsn{\overline{U}: U \in \fU_A}$.
\end{proof}