More housekeeping

This commit is contained in:
Bokuan Li
2026-10-29 16:24:08 -04:00
parent f0f13e8ed9
commit 714925d34b
2 changed files with 31 additions and 4 deletions
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\end{enumerate} \end{enumerate}
\end{proposition} \end{proposition}
\begin{proof} \begin{proof}
(1), (2): Let $D \subset E$ be a countable dense subset. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $S$ is embedded as a subspace of $K^D$. By \autoref{theorem:uniform-metrisable}, $\real^D$ is metrisable. By \autoref{proposition:separable-product}, $K^D$ is separable. Thus $S$ is also metrisable and separable by \autoref{proposition:separable-metric-space}. (1), (2): Let $D \subset E$ be a countable dense subset. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, the topology of pointwise convergence on $E$ and the topology of pointwise convergence on $E$ coincide on $S$, so $S$ is embedded as a subspace of $K^D$. By \autoref{theorem:uniform-metrisable}, $K^D$ is metrisable. By \autoref{proposition:separable-product}, $K^D$ is separable. Thus $S$ is also metrisable and separable by \autoref{proposition:separable-metric-space}.
(3): For any $A \subset E$, $A = \bigcup_{n \in \natp}A \cap nS$. By \autoref{proposition:separable-metric-space}, $A \cap nS$ is separable for each $n \in \natp$. Therefore $A$ is also separable. (3): For any $A \subset E$, $A = \bigcup_{n \in \natp}A \cap nS$. By \autoref{proposition:separable-metric-space}, $A \cap nS$ is separable for each $n \in \natp$. Therefore $A$ is also separable.
\end{proof} \end{proof}
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\end{proof} \end{proof}
\begin{example} \begin{example}
\label{example:open-mapping-counters} \label{example:open-mapping-counters}
@@ -160,3 +159,31 @@
\end{example} \end{example}
\begin{theorem}[Closed Complementation Theorem]
\label{theorem:closed-complement}
Let $E$ be a complete metric TVS, $\bracsn{0} \ne F \subsetneq E$ be a closed subspace, then the following are equivalent:
\begin{enumerate}
\item $F$ admits a closed complementary subspace.
\item There exists $P \in L(E; E)$ such that $P^2 = P$ and $P(E) = F$.
\end{enumerate}
\end{theorem}
\begin{proof}
(2) $\Rightarrow$ (1): Let $G = \ker(P)$, then the mapping
\[
E \to F \times G \quad x \mapsto (Px, (1 - P)x)
\]
is a toplinaer isomorphism.
(1) $\Rightarrow$ (2): Let $G$ be a closed complementary subspace of $E$, so that the mapping
\[
T: F \times G \to E \quad (x, y) \mapsto x + y
\]
is a continuous, bijective linear map. As $F, G \subset E$ are both closed, they are complete by \autoref{proposition:complete-closed}, and $E \times F$ is a complete metric TVS by \autoref{proposition:product-complete}. By the \hyperref[Open Mapping Theorem]{theorem:open-mapping}, the $T$ is a toplinear isomorphism. As such, $P$ is given by the following composition
\[
\xymatrix{
E \ar@{->}[r]^{T^{-1}} & F \times G \ar@{->}[r]^{\pi_1} & F
}
\]
\end{proof}