# Continuity proof

• Aug 4th 2013, 06:36 PM
krcmd
Continuity proof
Munkres p30 prob. 6: " $Let X = A \cup B$, where A and B are subspaces of X. Let $f:X \rightarrow Y$; suppose that the restricted functions $f|A:A \rightarrow Y$ and $f|B \rightarrow Y$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 x0 $\epsilon A$. By the definition of continuity, for every open set $V \subset Y$ containing f(x0) there is an open set Ya $\subset A$ containing x0 s.t. f|Ya $\subset V$. If x0 $\epsilon Ext B$, then Ya $\subset X$ satisfies the condition for continuity. Otherwise, x0 $\epsilon B$. In that case there is similarly an open set Yb containing x0 s.t. f|Yb $\subset V$. Choose Yx= Ya $\cap$Yb. The same reasoning applies with the sets A and B reversed. Therefore for every x0 $\epsilon (A \cup B)$ there is an open set (i.e. Yx) containing x0 s.t. f|Yx $\subset V$. Then f is continuous on X.
• Aug 5th 2013, 05:59 PM
johng
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.