
Continuity proof
Munkres p30 prob. 6: "$\displaystyle Let X = A \cup B$, where A and B are subspaces of X. Let $\displaystyle f:X \rightarrow Y$; suppose that the restricted functions $\displaystyle fA:A \rightarrow Y $ and $\displaystyle fB \rightarrow Y $are continuous. Show that if both A and B are closed in X, then f is continuous." Please let me know if the proof below is sound. I would also appreciate help on how to write it better. Thank you.
Assume x_{0}$\displaystyle \epsilon A$. By the definition of continuity, for every open set $\displaystyle V \subset Y$ containing f(x_{0}) there is an open set Y_{a}$\displaystyle \subset A$ containing x_{0} s.t. fY_{a}$\displaystyle \subset V$. If x_{0}$\displaystyle \epsilon Ext B$, then Y_{a}$\displaystyle \subset X$ satisfies the condition for continuity. Otherwise, x_{0}$\displaystyle \epsilon B$. In that case there is similarly an open set Y_{b} containing x_{0} s.t. fY_{b}$\displaystyle \subset V$. Choose Y_{x}= Y_{a}$\displaystyle \cap $Y_{b}. The same reasoning applies with the sets A and B reversed. Therefore for every x_{0}$\displaystyle \epsilon (A \cup B)$ there is an open set (i.e. Y_{x)} containing x_{0} s.t. fY_{x}$\displaystyle \subset V$. Then f is continuous on X.

Re: Continuity proof
Hi krcmd,
It seems to me that you are glossing over what "f restricted to A" continuous means. In particular the only way to make sense of this is to put the relative topology on A. Then of course an open sets of A is the intersection of an open set in X with A, similarly for a closed set of A.
I think the easiest proof is to show for F closed in Y, f^{1}(F) is closed in X. If you have question about proceeding in this direction, post your problem.