Added compact definition.
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Bokuan Li
2026-06-26 00:17:28 -04:00
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@@ -45,6 +45,12 @@
then $\fB$ is a filter base consisting of closed sets. By assumption, there exists $x \in \bigcap_{i \in I}U_j^c$, so $\mathbf{U}$ is not an open cover, contradiction.
\end{proof}
\begin{definition}[Relatively Compact]
\label{definition:relatively-compact}
Let $X$ be a topological space and $A \subset X$, then $A$ is \textbf{relatively compact} if $\ol A$ is compact.
\end{definition}
\begin{proposition}
\label{proposition:compact-extensions}
Let $X$ be a topological space and $E, F \subset X$ be compact, then the following sets are compact: