Proposition 2.18

Proposition 2.18: Let $$X$$ be a topological space and let $$A$$ be a subset of $$X$$. Then $$A$$ is closed iff $$A$$ contains all its limit points.

Ian_mi
First suppose $$A$$ is closed. Let $$p \in A^c$$. Then $$A^c$$ is an open set containing $$p$$ but no point of $$A$$. Therefore $$p$$ is not a limit point of $$A$$ so $$A$$ must contain all its limit points.

Conversely, suppose $$A$$ contains all its limit points. Let $$p \in A^c$$. Then $$p$$ is not a limit point of $$A$$ so there exists an open set $$U_p$$ such $$U_p \subseteq A^c$$. Let $$U = \bigcup_{p \in A^c} U_p$$. Then $$U$$ is the union of open sets and hence open and for every $$p \in A^c$$, $$p \in U_p \subseteq \bigcup_{p \in A^c} U_p = U$$ so $$A^c \subseteq U$$. In addition, $$U_p \subseteq A^c$$ so $$U = \bigcup_{p \in A^c} U_p \subseteq A^c$$. Thus $$U = A^c$$ so $$A$$ is closed.

Ian mi 05:11, March 15, 2011 (UTC)Ian_mi

'''Spot on. --Steven.clontz 20:36, March 17, 2011 (UTC)'''