Proposition 1.21

$$f: X \rightarrow Y$$ is continuous if and only if $$f^{-1}(U)$$ is open in $$X$$ for every open $$U \subset Y$$.

Ian_mi
First, suppose that $$ f: X \rightarrow Y $$ is continuous and let $$ U \subset Y $$ be open. Let $$ x \in f^{-1}(U) $$. Then $$ f(x) \in U $$ so $$ \exists \epsilon > 0 $$ such that $$B(f(x), \epsilon) \subset U$$. $$ f $$ is continuous so $$ \exists \delta > 0 $$ such that $$d(f(y),f(x)) < \epsilon$$ whenever $$ d(x,y) < \delta $$. In other words, if $$y \in B(x,\delta) $$ then $$ f(y) \in B(f(x), \epsilon) \subset U$$ so $$ y \in f^{-1}(U) $$. Thus $$ B(x,\delta) \subset f^{-1}(U) $$ so $$ f^{-1}(U) $$ is open.

Conversely, suppose that $$ f^{-1}(U) $$ is open whenever $$U$$ is open. Let $$x \in X$$ and let $$ \epsilon > 0 $$. Then $$B(f(x),\epsilon)$$ is open so $$f^{-1}(B(f(x),\epsilon))$$ is open. $$x \in f^{-1}(B(f(x),\epsilon))$$ so $$\exists \delta > 0$$ such that $$B(x, \delta) \subset f^{-1}(B(f(x),\epsilon))$$. In other words, if $$d(y,x) < \delta $$ then $$ y \in f^{-1}(B(f(x),\epsilon)) $$ and hence $$f(y) \in B(f(x),\epsilon)$$. Thus $$d(f(y),f(x)) < \epsilon$$ so f is continuous.

Ian mi 21:57, February 27, 2011 (UTC)ian_mi

'''Looks good to me. --Steven.clontz 04:15, March 14, 2011 (UTC)'''