# Characterization of compact spaces

Theorem: Let $$\tau$$ be a topology on $$X$$ and let $$\mathcal{B}$$ be a base for $$\tau$$ on $$X$$.

$$(X,\tau) \text{ is compact space}$$

$$\Leftrightarrow$$

$$(\forall\mathcal{A}\subseteq\mathcal{B})[X=\cup\mathcal{A}\to (\exists\mathcal{A}^*\subseteq\mathcal{A})(|\mathcal{A}^*|<\aleph_0)(X=\cup\mathcal{A}^*)].$$

Proof: $$(\Rightarrow):$$ Let $$\mathcal{A}\subseteq \mathcal{B}$$ and $$X=\cup\mathcal{A}.$$ $$\left.\begin{array}{rr}(\mathcal{A}\subseteq \mathcal{B})(X=\cup\mathcal{A}) \\ \\ \mathcal{B} \ \text{is a base for } \tau \Rightarrow \mathcal{B}\subseteq\tau \end{array} \right\}\Rightarrow \begin{array}{rr} \\ \\ \left. \begin{array}{cc} (\mathcal{A}\subseteq \tau)(X=\cup\mathcal{A}) \\ \\ (X,\tau) \text{ is a compact space}\end{array} \right\} \Rightarrow \end{array}$$

$$\Rightarrow (\exists\mathcal{A}^*\subseteq\mathcal{A})(|\mathcal{A}^*|<\aleph_0)(X=\cup\mathcal{A}^*).$$

$$\mbox{}$$

$$(\Leftarrow):$$ Let $$\mathcal{A}\subseteq \tau$$ and $$X=\cup\mathcal{A}.$$ $$\left.\begin{array}{rr} (A\in \mathcal{A}\subseteq \tau)(X=\cup\mathcal{A}) \\ \\ \mathcal{B} \ \text{ is a base for } \tau \end{array} \right\}\Rightarrow \begin{array}{rr} \\ \\ \left. \begin{array}{rr} (\exists \mathcal{B}_A\subseteq \mathcal{B})(A=\cup\mathcal{B}_A)(X=\cup_{A\in\mathcal{A}}(\cup\mathcal{B}_A)) \\ \\ \mathcal{A}^*:=\cup\{\mathcal{B}_A|A\in\mathcal{A}\}\end{array} \right\} \Rightarrow \end{array}$$

$$\left.\begin{array}{rr}\Rightarrow (\mathcal{A}^*\subseteq\mathcal{B})(X=\cup\mathcal{A}^*) \\ \\ \text{Hypothesis}\end{array}\right\}\Rightarrow (\exists\mathcal{A}^{**}\subseteq\mathcal{A})(|\mathcal{A}^{**}|<\aleph_0)(X=\cup\mathcal{A}^{**}).$$

Is there any problem in proof which is given above?

• How do you define the $\mathcal{A}^{\ast\ast}$ in the last part from $\mathcal{A}^\ast$? – Henno Brandsma Dec 29 '18 at 22:37
• I can’t read all this logic symbolism. – Randall Dec 30 '18 at 1:34
• Proofs are meant to be read by humans, not machines! – Henno Brandsma Dec 30 '18 at 9:10
• Dear Henno. I also think as you. There is a problem in the proof as you've been emphasizing. – murad.ozkoc Dec 30 '18 at 9:34

The right to left implication can also be done slightly differently: let $$\mathcal{O}$$ be any open cover of $$X$$ and for each $$x \in X$$ we pick some $$O_x \in \mathcal{O}$$ such that $$x \in O_x$$, and then (as we have a base) a $$B_x \in \mathcal{B}$$ such that $$x \in B_x \subseteq O_x$$. Finitely many $$B_x$$ cover $$X$$ by the assumption on basic covers and the corresponding $$O_x$$ then cover $$X$$ too (as they're larger).