Proposition 2.39

Ian_mi
First, suppose $$\{x_n\}$$ converges to $$x$$. Let $$G$$ be an open set containing $$x$$. Then $$\exists N \in \mathbb{N}$$ such that $$x_n \in G$$ for all $$n \ge N$$. Therefore if $$x_n \notin G$$ then $$n \in \{1, \ldots, N-1\}$$ which is finite so $$x_n \in G$$ for all but finitely many $$n$$.

Second, suppose that for any open set $$G$$ containing $$x$$, $$x_n \in G$$ for all but finitely many $$n$$. Let $$G$$ be an open set containing $$x$$. Then there exists a finite set $$\mathcal{N}$$ such that $$x_n \in G$$ for all $$n \notin \mathcal{N}$$. $$\mathcal{N}$$ is finite so it has a maximal element. Let $$N = \max\{\mathcal{N}\} + 1$$. Then for $$n \ge N$$, $$n \notin \mathcal{N}$$ so $$x_n \in G$$. Thus $$\{x_n\}$$ converges to $$x$$.

Ian mi 05:21, March 24, 2011 (UTC)Ian_mi