Proposition 2.22

Proposition 2.22: Let $$A \subseteq X$$, and $$\mathcal{C}(A)= \{ C : C \text{ is closed and } C \supseteq A\}$$. Then $$\overline{A}= \bigcap \mathcal{C}(A)$$, that is, $$\overline{A}$$ is the intersection of all closed supersets of $$A$$.

Ian_mi
Let $$F$$ be a closed set containing A and let $$x \in \overline{A}$$. If $$x \in A$$ then $$x \in F$$ since $$A \subseteq F$$. Otherwise $$x \in A'$$ so x is a limit point of $$A$$. Therefore every open set containing $$x$$ contains a point of $$A$$ and hence of $$F$$ which is distinct from $$x$$. Thus $$x$$ is a limit point of $$F$$ so $$x \in F$$ since $$F$$ is closed. Therefore $$\overline{A} \subseteq F$$ and hence $$\overline{A} \subseteq \bigcap \{F : A \subseteq F \text{ and } F \text{ is closed}\}$$.

Let $$x$$ be a limit point of $$\overline{A}$$ such that $$x \notin A$$ and let $$G$$ be an open set containing $$x$$. Then $$G$$ contains a point $$p$$ such that $$p \in \overline{A}$$. If $$p \notin A$$ then $$p \in A'$$ so $$p$$ is a limit point of $$A$$. Therefore there exists a point $$q \in G$$ such that $$q \in A$$. In either case $$G$$ contains a point of $$A$$ and since $$x \notin A$$ this point is distinct from $$x$$. Therefore $$x$$ is a limit point of $$A$$ and hence $$x \in A' \subseteq \overline{A}$$. If $$x$$ is any other limit point of $$\overline{A}$$ then $$x \in A \subseteq \overline{A}$$. Thus $$\overline{A}$$ contains all of its limit points so $$\overline{A}$$ is closed. $$A \subseteq \overline{A}$$ so $$\overline{A} \subseteq \bigcap \{F : A \subseteq F \text{ and } F \text{ is closed}\}$$. Thus $$\overline{A} = \bigcap \{F : A \subseteq F \text{ and } F \text{ is closed}\}$$.

Ian mi 04:38, March 22, 2011 (UTC)Ian_mi

Comments
@lan_mi

At the end of the first paragraph, did you mean "Abar in F" instead of "A in F"? Same for the next equation? Tony Bruguier 16:29, March 22, 2011 (UTC)

Yes, thanks.