More housekeeping
This commit is contained in:
@@ -12,7 +12,7 @@
|
||||
\end{enumerate}
|
||||
\end{proposition}
|
||||
\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.
|
||||
\end{proof}
|
||||
|
||||
@@ -150,7 +150,6 @@
|
||||
|
||||
\end{proof}
|
||||
|
||||
|
||||
\begin{example}
|
||||
\label{example:open-mapping-counters}
|
||||
|
||||
@@ -160,3 +159,31 @@
|
||||
\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}
|
||||
|
||||
Reference in New Issue
Block a user