Let X be a topological space and let be a sequence of continuous functions from X to the real numbers. If f(x)= for all x in X, show that the set of points where f is continuous is a set.
Let
If then , so is open. Also, f is continuous at x if and only if . Thus the set of points of continuity of f is a .
What bothers me about that argument is that it seems to apply to an arbitrary function f. (It doesn't assume that f is the pointwise limit of a sequence of continuous functions.) Am I missing something?
I think you're right.
But this problem (and the suggested solution) also reminds me of something else, which is somewhat more interesting: under the additional hypothesis that is a complete metric space (or any Baire space), is continuous on a dense set.
This can be proved by considering . This is a closed subset of . Then let . This is an open subset, and it is dense because of Baire's theorem: if not, then there would be and a closed ball such that , which means for all , hence , and Baire's theorem would say , in contradiction with (which holds because the pointwise convergence gives ). Finally, Baire's theorem shows that is dense. And one can see that is continuous on .