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Bokuan Li
2026-03-06 14:06:15 -05:00
parent 173727665b
commit 5034bc4220
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34
src/fa/tvs/definition.tex
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@@ -31,6 +31,7 @@
\[
U = \bracs{(x + z, y + z)|(x, y) \in U}
\]
and $\fU$ is \textbf{translation-invariant} if there exists a fundamental system of translation-invariant entourages.
\end{definition}
@@ -39,7 +40,7 @@
Let $E$ be a vector space and $\fU$ be a translation-invariant uniformity, then $\fU$ admits a fundamental system of symmetric, translation-invariant entourages.
\end{lemma}
\begin{proof}
Let $z \in E$, then the map $(x, y) \mapsto (x + z, y + z)$ is a bijection. Thus for any translation-invariant entourages $U, V \in \fU$, $(U \cap V) + z = (U + z) \cap (V + z)$, and $U \cap V$ is translation-invariant. By \ref{lemma:symmetricfundamentalentourage}, $\fU$ admits a fundamental system of symmetric, translation-invariant entourages.
Let $z \in E$, then the map $(x, y) \mapsto (x + z, y + z)$ is a bijection. Thus for any translation-invariant entourages $U, V \in \fU$, $(U \cap V) + z = (U + z) \cap (V + z)$, and $U \cap V$ is translation-invariant. By \autoref{lemma:symmetricfundamentalentourage}, $\fU$ admits a fundamental system of symmetric, translation-invariant entourages.
\end{proof}
@@ -60,13 +61,13 @@
\item[(UB2)] Let $V \in \fB_0$, then by (TVS1), there exists $W \in \fB_0$ such that $W + W \subset V$. For any $(x, y), (y, z) \in U_W$, $(x - y), (y - z) \in W$, so $(x - z) \in V$. Thus $U_W \circ U_W \subset U_V$.
\end{enumerate}
By \ref{proposition:fundamental-entourage-criterion}, $\fB$ forms a fundamental system of entourages for a translation-invariant uniformity $\fU$ on $E$.
By \autoref{proposition:fundamental-entourage-criterion}, $\fB$ forms a fundamental system of entourages for a translation-invariant uniformity $\fU$ on $E$.
(1): Let $\mathfrak{V}$ be a translation-invariant uniformity on $E$ inducing the topology. For any symmetric, translation-invariant entourage $V \in \mathfrak{V}$, $V(x) = V(0) + x$ for all $x \in E$, and $(x, y) \in V$ if and only if $y - x \in V$, if and only if $x - y \in V$. Thus $V = U_{V(0)}$.
Let $W \in \cn(0)$, then by \ref{lemma:translation-invariant-symmetric}, there exists a symmetric, translation-invariant entourage $V \in \mathfrak{V}$ such that $V(0) \subset W$, and $V \subset U_W$. Thus $\mathfrak{V} \supset \fU$.
Let $W \in \cn(0)$, then by \autoref{lemma:translation-invariant-symmetric}, there exists a symmetric, translation-invariant entourage $V \in \mathfrak{V}$ such that $V(0) \subset W$, and $V \subset U_W$. Thus $\mathfrak{V} \supset \fU$.
Let $V \in \mathfrak{V}$. Using \ref{lemma:translation-invariant-symmetric}, assume without loss of generality that $V$ is symmetric and translation-invariant, then there exists $W \in \fB_0$ with $W \subset V(0)$. In which case, $U_W \subset V$, and $\fU \supset \mathfrak{V}$.
Let $V \in \mathfrak{V}$. Using \autoref{lemma:translation-invariant-symmetric}, assume without loss of generality that $V$ is symmetric and translation-invariant, then there exists $W \in \fB_0$ with $W \subset V(0)$. In which case, $U_W \subset V$, and $\fU \supset \mathfrak{V}$.
\end{proof}
\begin{proposition}
@@ -75,14 +76,16 @@
\[
\ol{A} = \bigcap_{U \in \fB}\bracs{A + U| U \in \fB}
\]
\end{proposition}
\begin{proof}
Let $V \in \cn(0)$ be balanced and $U_V = \bracs{(x, y) \in E \times E| x - y \in V}$, then $y \in U_V(A)$ if and only if there exists $x \in A$ such that $(x, y) \in U_V$. This is equivalent to $x - y \in V$ and $y - x \in V$, so $U_V(A) = A + V$.
Assume without loss of generality that $\fB$ consists of symmetric entourages. By \ref{proposition:tvs-uniform}, $\bracs{U_V|V \in \fB}$ forms a fundamental system of entourages for $E$, and \ref{proposition:uniformclosure} implies that
Assume without loss of generality that $\fB$ consists of symmetric entourages. By \autoref{proposition:tvs-uniform}, $\bracs{U_V|V \in \fB}$ forms a fundamental system of entourages for $E$, and \autoref{proposition:uniformclosure} implies that
\[
\ol{A} = \bigcap_{V \in \fB}\bracs{U_V(A)| U \in \fB} = \bigcap_{V \in \fB}U + A
\]
\end{proof}
\begin{proposition}[{{\cite[I.1.1]{SchaeferWolff}}}]
@@ -94,10 +97,11 @@
\end{enumerate}
\end{proposition}
\begin{proof}
$(1)$: For every $x \in B$, $A + x$ is open by \ref{definition:translation-invariant-topology}, so
$(1)$: For every $x \in B$, $A + x$ is open by \autoref{definition:translation-invariant-topology}, so
\[
A + B = \bigcup_{x \in B}(A + x)
\]
is open.
$(2)$: Let $x \in \overline{A + B}$, then there exists a filter $\fF \subset 2^{A \cup B}$ converging to $x$. For any $U \in \fF$, $U \cap (A + B) \ne \emptyset$, so $(U - B) \cap A \ne \emptyset$, and $\fB = \bracs{U - B| U \in \fF}$ is a filter base in $A$. By compactness of $A$, there exists $y \in A$ such that
@@ -105,14 +109,17 @@
y \in \bigcap_{U \in \fF}\overline{U - B} = \bigcap_{U \in \fF}\overline{(-B) + U}
\]
By \ref{proposition:tvs-closure}, $\overline{(-B) + U} \subset (-B) + U + U$, so
By \autoref{proposition:tvs-closure}, $\overline{(-B) + U} \subset (-B) + U + U$, so
\[
y \in \bigcap_{U \in \fF}\overline{(-B) + U} \subset \bigcap_{U \in \fF}[(-B) + U + U]
\]
Since $\fF$ converges to $x$, (TVS1) implies that $\bracs{U + U| U \in \fF}$ contains a neighbourhood base of $x$. Thus
\[
y \in \bigcap_{U \in \fF}[(-B) + U + U] \subset \bigcap_{V \in \cn(0)}[(x-B) + V] = \overline{x - B} = x - B
\]
so $x \in y + B \subset A + B$.
\end{proof}
@@ -137,12 +144,13 @@
\begin{proof}
Firstly, (TVS2) implies that every neighbourhood of $0$ is circled.
By \ref{proposition:uniform-neighbourhoods}, $E$ admits a fundamental system of neighbourhoods consisting of open sets or closed sets.
By \autoref{proposition:uniform-neighbourhoods}, $E$ admits a fundamental system of neighbourhoods consisting of open sets or closed sets.
Let $U \in \cn^o(0)$ be open. By (TVS2), there exists $r > 0$ such that $\lambda U \subset U$ for all $\lambda \in K$ with $\abs{\lambda} \le r$. Define
\[
V = \bigcup_{\substack{\lambda \in K \\ \abs{\lambda} \le r}} \lambda U \subset U
\]
then for any $x \in V$, there exists $\lambda \in K$ with $\abs{\lambda} \le r$ and $y \in U$ such that $x = \lambda y$. In which case, for any $\mu \in K$ with $\abs{\mu} \le 1$, $\mu(\lambda y) = (\mu \lambda) y \in \mu\lambda U$. Since $\abs{\mu \lambda} \le r$, $\mu \lambda U \subset V$. Thus $V \subset U$ is balanced.
Let $U \in \cn(0)$ be closed, then there exists a balanced neighbourhood $V \in \cn^o(0)$ such that $V \subset U$. In which case, for any $\lambda \in K$ with $0 < \abs{\lambda} \le 1$, $\lambda \overline{V} = \overline{\lambda V} \subset \overline{V}$ by (TVS2). Therefore $\overline{V} \subset U$ is balanced as well.
@@ -166,19 +174,20 @@
\end{enumerate}
\end{proposition}
\begin{proof}
\textbf{Forward:} By \ref{proposition:tvs-good-neighbourhood-base}, there exists a fundamental system of neighbourhoods $\fB \subset \cn_E(0)$ consisting of circled and radial sets. By (TVS1), $\fB$ satisfies (TVB1).
\textbf{Forward:} By \autoref{proposition:tvs-good-neighbourhood-base}, there exists a fundamental system of neighbourhoods $\fB \subset \cn_E(0)$ consisting of circled and radial sets. By (TVS1), $\fB$ satisfies (TVB1).
\textbf{Converse:} For each $V \in \fB$, let $U_V = \bracs{(x, y) \in E|x - y \in V}$, then $U_V$ is symmetric and translation-invariant by (TVB1). Let
\[
\mathfrak{V} = \bracs{U_V|V \in \fB}
\]
then
\begin{enumerate}
\item[(FB1)] For any $V, V' \in \fB$, there exists $W \in \fB$ with $W \subset V \cap V'$. In which case, $U_{V} \cap U_{V'} \supset U_W \in \mathfrak{V}$.
\item[(UB1)] For any $x \in E$ and $V \in \fB$, $x - x = 0 \in V$, so $\Delta \subset U_V$.
\item[(UB2)] For any $V \in \fB$, by (TVB1), there exists $W \in \fB$ such that $W + W \subset V$. In which case, for any $x, y, z \in E$ with $x - y, y - z \in W$, $x - z \in V$. Therefore $U_W \circ U_W \subset U_V$.
\end{enumerate}
By \ref{proposition:fundamental-entourage-criterion}, there exists a unique uniformity $\fU$ on $E$ for which $\mathfrak{V}$ is a fundamental system of entourages.
By \autoref{proposition:fundamental-entourage-criterion}, there exists a unique uniformity $\fU$ on $E$ for which $\mathfrak{V}$ is a fundamental system of entourages.
(1): Since $\mathfrak{V}$ is translation-invariant, so is $\fU$.
@@ -197,10 +206,11 @@
\lambda x - \lambda' x' \in \lambda V + \eps(\abs \mu + 1)V \subset \abs{\lambda} V + \eps(\abs \mu + 1)V
\]
Let $W \in \fB$ and $\eps > 0$ such that $\eps(\abs{\mu}+1) \le 1$, then by repeated application of (TVB1), there exists $V \in \fB$ such that $\abs{\lambda} V + V \subset W$. Therefore scalar multiplication is jointly continuous.
\end{enumerate}
(Uniqueness): Let $\mathcal{S} \subset 2^E$ be a topology on $E$ satisfying (1) and (2), then for each $x \in E$, $\cn_{(E, \mathcal{S})}(x) = \cn_{(E, \mathcal{T})}(x)$. By \ref{proposition:neighbourhoodcharacteristic}, $\mathcal{S} = \mathcal{T}$.
(Uniqueness): Let $\mathcal{S} \subset 2^E$ be a topology on $E$ satisfying (1) and (2), then for each $x \in E$, $\cn_{(E, \mathcal{S})}(x) = \cn_{(E, \mathcal{T})}(x)$. By \autoref{proposition:neighbourhoodcharacteristic}, $\mathcal{S} = \mathcal{T}$.
\end{proof}
@@ -209,5 +219,5 @@
Let $E$ be a TVS over $K \in \RC$, then $E$ is locally connected.
\end{proposition}
\begin{proof}
Let $U \in \cn(0)$ be radial, then for any $y \in U$, the mapping $t \mapsto ty$ is a path from $0$ to $y$ contained in $U$. Thus $U$ is path-connected. By \ref{proposition:tvs-0-neighbourhood-base}, the radial neighbourhoods of $0$ forms a fundamental system of neighbourhoods, so the path-connected neighbourhoods of $0$ forms a fundamental system as well.
Let $U \in \cn(0)$ be radial, then for any $y \in U$, the mapping $t \mapsto ty$ is a path from $0$ to $y$ contained in $U$. Thus $U$ is path-connected. By \autoref{proposition:tvs-0-neighbourhood-base}, the radial neighbourhoods of $0$ forms a fundamental system of neighbourhoods, so the path-connected neighbourhoods of $0$ forms a fundamental system as well.
\end{proof}