Proving arbitrary intersection of closed sets from closure axioms

general-topology

I am given the following axioms to define the closure of a sub-set $A$ in $X$:

  1. $\overline \emptyset$ = $\emptyset$
  2. $\overline X$ = $X$
  3. $A \subset \overline A$
  4. $\overline {A \cup B} = \overline A \cup \overline B$
  5. $\overline{\overline A} = \overline A$

From these I want to show that the definition of a closed set as a set $A = \overline A$ satisfies the axioms of a closed set in a topological space. But I am stumped on the axiom of arbitrary intersection:

Given a family $\{F_\alpha\}_{\alpha \in I}$ of closed sets $\bigcap_{\alpha \in I} F_\alpha$ is closed.

Best Answer

Let $F=\bigcap_{\alpha\in I}F_\alpha$. For each $\alpha\in I$ we have $F\subseteq F_\alpha$, so by (4)

$$\overline{F}\subseteq\overline{F}\cup\overline{F_\alpha}=\overline{F\cup F_\alpha}=\overline{F_\alpha}\,,$$

and therefore

$$F\subseteq\overline{F}\subseteq\bigcap_{\alpha\in I}\overline{F_\alpha}=\bigcap_{\alpha\in I}F_\alpha=F\,,$$

i.e., $F=\overline{F}$.

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