Added bornological spaces.

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Bokuan Li
2026-05-02 15:59:03 -04:00
parent dcf11fb978
commit 1e53581113
8 changed files with 82 additions and 22 deletions

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@@ -3,13 +3,13 @@
\begin{definition}[Bounded]
\label{definition:bounded}
Let $E$ be a TVS over $K \in \RC$ and $B \subset E$, then the following are equivalent:
Let $(E, \topo)$ be a TVS over $K \in \RC$ and $B \subset E$, then the following are equivalent:
\begin{enumerate}
\item For every $U \in \cn(0)$, there exists $\lambda \in K$ such that $\lambda U \supset B$.
\item For every $\seq{x_n} \subset B$ and $\seq{\lambda_n} \subset K$ such that $\lambda_n \to 0$, $\lambda_n x_n \to 0$ as $n \to \infty$.
\end{enumerate}
If the above holds, then $B$ is \textbf{bounded}. The collection $B(E)$ is the set of all bounded sets of $E$.
If the above holds, then $B$ is \textbf{bounded}. The collection $B(E) = B(E, \topo)$ is the set of all bounded sets of $E$.
\end{definition}
\begin{proof}
(1) $\Rightarrow$ (2): Let $U \in \cn_E(0)$ be circled, then there exists $k \in \natp$ such that $kU \supset B$. Since $\lambda_n \to 0$ as $n \to \infty$, there exists $N \in \natp$ such that $|\lambda_n| \le 1/k$ for all $n \ge N$. In which case, $\lambda_n x_n \in \lambda_n B \subset U$ for all $n \ge N$.