From 220c263f8f9212f95ca541294e23171221689f7d Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Fri, 23 Jan 2026 16:44:29 -0500 Subject: [PATCH] Added quotient spaces. --- src/fa/tvs/definition.tex | 24 ++++++++++++++----- src/fa/tvs/index.tex | 1 + src/fa/tvs/quotient.tex | 49 +++++++++++++++++++++++++++++++++++++++ 3 files changed, 68 insertions(+), 6 deletions(-) create mode 100644 src/fa/tvs/quotient.tex diff --git a/src/fa/tvs/definition.tex b/src/fa/tvs/definition.tex index 09eb15b..4291160 100644 --- a/src/fa/tvs/definition.tex +++ b/src/fa/tvs/definition.tex @@ -150,17 +150,25 @@ \begin{proposition}[{{\cite[1.2]{SchaeferWolff}}}] \label{proposition:tvs-0-neighbourhood-base} - Let $E$ be a vector space of $K \in \RC$, and $\topo$ be a vector space topology on $E$, then there exists a fundamental system of neighbourhoods $\fB \subset \cn_E(0)$ such that: + Let $E$ be a vector space over $K \in \RC$, and $\topo$ be a vector space topology on $E$, then there exists a fundamental system of neighbourhoods $\fB \subset \cn_E(0)$ such that: \begin{enumerate} \item[(TVB1)] For each $U \in \fB$, there exists $V \in \fB$ such that $V + V \subset U$. \item[(TVB2)] For each $U \in \fB$, $U$ is circled and radial. \end{enumerate} - Conversely, if $\fB \subset 2^E$ is a family of sets that contain $0$ and satisfies (TVB1) and (TVB2), then there exists a unique topology $\topo$ on $E$ for which $\fB$ is a fundamental system of neighbourhoods at $0$. + Conversely, if $\fB \subset 2^E$ is a family of sets that contain $0$ and satisfies (TVB1) and (TVB2), then there exists a unique topology $\topo$ on $E$ such that: + \begin{enumerate} + \item $\topo$ is translation-invariant. + \item $\fB$ is a fundamental system of neighbourhoods at $0$ for $\topo$. + \end{enumerate} + Moreover, + \begin{enumerate} + \item[(3)] $(E, \topo)$ is a TVS. + \end{enumerate} \end{proposition} \begin{proof} - (1): 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 \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). - (2): 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 + \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 \[ \mathfrak{V} = \bracs{U_V|V \in \fB} \] @@ -172,7 +180,11 @@ \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. - It remains to verify that $\mathfrak{V}$ induces a vector space topology on $E$. + (1): Since $\mathfrak{V}$ is translation-invariant, so is $\fU$. + + (2): By definition of the uniform topology, $\fB = \bracs{U_V(0)|V \in \fB}$ is a fundamental system of neighbourhoods at $0$. + + (3): \begin{enumerate} \item[(TVS1)] Let $V \in \fB$, then there exists $W \in \fB$ such that $W + W \subset V$ by (TVB1). In which case, for any $x, x', y, y'$ with $x - x' \in W$ and $y - y' \in W$, $(x + y) - (x' + y') \in W + W \subset V$. \item[(TVS2)] Let $V \in \fB$, $\eps > 0$, $x, x' \in E$ with $x - x' \in V$, and $\lambda, \lambda' \in K$ with $\abs{\lambda - \lambda'} < \eps$, then @@ -188,5 +200,5 @@ 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} - Finally, by definition of the uniform topology, $\fB = \bracs{U_V(0)|V \in \fB}$ is a fundamental system of neighbourhoods at $0$. + (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}, $\mathcal{S} = \mathcal{T}$. \end{proof} diff --git a/src/fa/tvs/index.tex b/src/fa/tvs/index.tex index 42827b5..e52afc8 100644 --- a/src/fa/tvs/index.tex +++ b/src/fa/tvs/index.tex @@ -5,5 +5,6 @@ \input{./src/fa/tvs/bounded.tex} \input{./src/fa/tvs/dual.tex} \input{./src/fa/tvs/continuous.tex} +\input{./src/fa/tvs/quotient.tex} \input{./src/fa/tvs/completion.tex} \input{./src/fa/tvs/spaces-of-linear.tex} diff --git a/src/fa/tvs/quotient.tex b/src/fa/tvs/quotient.tex new file mode 100644 index 0000000..caeb69f --- /dev/null +++ b/src/fa/tvs/quotient.tex @@ -0,0 +1,49 @@ +\section{Quotient Spaces} +\label{section:tvs-quotient} + +\begin{definition}[Quotient TVS] +\label{definition:tvs-quotient} + Let $E$ be a TVS over $K \in \RC$, and $M \subset E$ be a vector subspace, then there exists $(\td E, \pi)$ such that: + \begin{enumerate} + \item $\td E$ is a TVS over $K \in \RC$. + \item $\pi \in L(E; \td E)$. + \item $\ker \pi \supset M$. + \item[(U)] For any topological space $F$ and $f \in C(E;F)$ such that $f(x) = f(y)$ whenever $x - y \in M$, there exists a unique $\td f \in C(\td E; F)$ such that the following diagram commutes + \[ + \xymatrix{ + E \ar@{->}[rd]^{f} \ar@{->}[d]_{\pi} & \\ + \widetilde E \ar@{->}[r]_{\tilde f} & F + } + \] + If $F$ is a TVS over $K$ and $f \in L(E; F)$, then $\td f \in L(E; F)$. + \end{enumerate} + The space $\td E = E/M$ is the \textbf{quotient} of $E$ by $M$. +\end{definition} +\begin{proof} + Let $\td E = E/M$ be the algebraic quotient of $E$ by $M$, and equip it with the quotient topology by $\pi$. + + (1): By \ref{definition:quotient-topology}, for each $\pi(U) \subset E/M$, $\pi(U)$ is open if and only if $U$ is open. Since the topology on $E$ is translation-invariant, so is the quotient topology on $E/M$. Let + \[ + \fB = \bracs{\pi(U)| U \in \cn(0) \text{ circled and radial}} + \] + Since the circled and radial neighbourhoods of $0$ forms a fundamental system of neighbourhoods at $0$ by \ref{proposition:tvs-good-neighbourhood-base}, $\fB$ is a fundamental system of neighbourhoods at $0$ for the quotient topology on $E/M$. In addition, + \begin{enumerate} + \item[(TVB1)] Let $U \in \cn(0)$ be circled and radial. For any $\lambda \in K$ with $\abs{\lambda} \le 1$, $\lambda \pi(U) = \pi(\lambda U) \subset \pi(U)$, so $\pi(U)$ is also circled. For any $x + M \in E/M$, there exists $\lambda \in K$ such that $x \in \lambda U$. In which case, $x \in \lambda U + M = \pi(U)$, so $\pi(U)$ is also radial. + \item[(TVB2)] For any $U \in \cn(0)$ circled and radial, by \ref{proposition:tvs-good-neighbourhood-base}, there exists $W \in \cn(0)$ such that $W + W \subset U$. In which case, $\pi(W) + \pi(W) \subset \pi(U)$. + \end{enumerate} + By \ref{proposition:tvs-0-neighbourhood-base}, there exists a unique translation-invariant topology on $E/M$ such that $\fB$ is a fundamental system of neighbourhoods at $0$, which must be the quotient topology on $E/M$. In which case, the quotient topology is a vector space topology by (3) of \ref{proposition:tvs-0-neighbourhood-base}. + + (2), (3), (U): By \ref{definition:quotient-topology}. +\end{proof} + +\begin{proposition} +\label{proposition:tvs-quotient-hausdorff} + Let $E$ be a TVS over $K \in \RC$, $M \subset E$ be a subspace, then $E/M$ is Hausdorff if and only if $M$ is closed. +\end{proposition} +\begin{proof} + The space $M$ is closed if and only if + \[ + M = \bigcap_{V \in \cn(0)}M + V + \] + which is equivalent to $E/M$ being Hausdorff. +\end{proof}