I'm reading Intro to Topology from Mendelson.
The entire problem statement is,
Prove that $X$ is compact if and only if for each family $\lbrace F_\alpha\rbrace_{\alpha\in I}$ of closed subsets of $X$ that has the finite intersection property, we have $\bigcap_{\alpha\in I} F_\alpha\neq\varnothing$.
My attempt at the proof is as follows:
First assume that $X$ is compact. For the sake of contradiction, suppose that there exists a family $\{ F_\alpha\}_{\alpha\in I}$ of closed subsets of $X$ with FIP such that $\bigcap_{\alpha\in I} F_\alpha=\varnothing$. Since $X$ is compact if $\lbrace F_\alpha\rbrace_{\alpha\in I}$ any family of closed sets such that $\bigcap_{\alpha\in I} F_\alpha=\varnothing$, then there exists a finite set of indices $\lbrace \alpha_1,\dots,\alpha_n\rbrace$ such that $\bigcap\limits_{i=1}^n F_{\alpha_i}=\varnothing$. Yet, since $\lbrace F_\alpha\rbrace_{\alpha\in I}$ has the finite intersection property, for every finite index of $J\subset I$, $\bigcap\limits_{\alpha\in I} F_\alpha\neq\varnothing$.
Suppose now that $X$ is not compact, that is, there exists an open cover $\lbrace U_\alpha\rbrace_{\alpha\in I}$ of $X$ with no finite subcover. Consider the set $F_\alpha=C(U_\alpha).$ Then $\lbrace F_\alpha\rbrace_{\alpha\in I}$ is a collection of closed subsets of $X$ with the finite intersection property and were $\bigcap_{\alpha\in I} F_\alpha=\varnothing.$
Thanks for any feedback!
\{\}
does the same thing as\lbrace\rbrace
with fewer characters, and that\bigcap
is usually more readable with subscripts/superscripts than is\cap
. The last sentence of the first paragraph is still completely unreadable. $\endgroup$