Used "separated" instead of Hausdorff in the context of topological vector spaces.

This commit is contained in:
Bokuan Li
2026-05-01 13:32:08 -04:00
parent caf7790b15
commit 3077563278
6 changed files with 8 additions and 8 deletions

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@@ -5,7 +5,7 @@
\label{definition:tvs-completion}
Let $E$ be a TVS over $K \in \RC$, then there exists $(\wh E, \iota)$ such that:
\begin{enumerate}
\item $\wh E$ is a complete Hausdorff TVS.
\item $\wh E$ is a complete separated TVS.
\item $\iota \in L(E; \wh E)$.
\item[(U)] For any $(F, T)$ satisfying (1) and (2), there exists a unique $\ol{T} \in L(\wh E; F)$ such that the following diagram commutes:
\end{enumerate}

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@@ -54,7 +54,7 @@
\begin{theorem}[Linear Extension Theorem (TVS)]
\label{theorem:linear-extension-theorem-tvs}
Let $E$ be a TVS over $K \in \RC$, $F$ be a complete Hausdorff TVS over $K$, $D \subset E$ be a dense subspace, and $T \in L(D; F)$, then:
Let $E$ be a TVS over $K \in \RC$, $F$ be a complete separated TVS over $K$, $D \subset E$ be a dense subspace, and $T \in L(D; F)$, then:
\begin{enumerate}
\item There exists an extension $\ol T \in L(E; F)$ such that $\ol T|_D = T$.
\item[(U)] For any $S \in C(E; F)$ satisfying (1), $S = \ol T$.

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@@ -40,7 +40,7 @@
\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.
Let $E$ be a TVS over $K \in \RC$, $M \subset E$ be a subspace, then $E/M$ is separated if and only if $M$ is closed.
\end{proposition}
\begin{proof}
The space $M$ is closed if and only if
@@ -48,5 +48,5 @@
M = \bigcap_{V \in \cn(0)}M + V
\]
which is equivalent to $E/M$ being Hausdorff.
which is equivalent to $E/M$ being separated.
\end{proof}