Proposition 2.41

''' Let $$X,Y$$ be topological spaces. If $$f: X \rightarrow Y$$ is continuous and $$\{x_n\} \subset X$$ converges to $$x \in X$$, then $$\{f(x_n)\}$$ converges to $$f(x) \in Y$$.

Olomana
This follows from the definitions of continuity and convergence.

Let $$V$$ be any open subset of $$Y$$ containing $$f(x)$$. Since $$f$$ is continuous, there is an open subset $$U$$ of $$X$$ containing $$x$$ such that $$f(U) \subseteq V$$.

Since $$U$$ is open and contains $$x$$, and since $$\{x_n\}$$ converges to $$x$$, there must be some $$N$$ such that $$x_n \in U$$ for every $$n > N$$.

Since $$f(U) \subseteq V$$, $$f(x_n) \in V$$ for every $$n > N$$.

Thus, for every open subset $$V$$ of $$Y$$ containing $$f(x)$$, there is an $$N$$ such that $$f(x_n) \in V$$ for every $$n > N$$. In other words, $$\{f(x_n)\}$$ converges to $$f(x)$$.

Olomana 17:18, March 22, 2011 (UTC)