1. Oct 7, 2004

### fourier jr

"Let f be a real-valued function defined for all real numbers. Prove that the set of points at which f is continuous is a $$G_\delta$$."
(a set is a $$G_\delta$$ if it is the intersection of a countable collection of open sets)

I think it's obvious that I should use the topological/open-set definition of continuous, and then intersect a bunch of open sets but I'm not sure how to write it down. (if that makes any sense)

2. Oct 7, 2004

### shmoe

Ahh, question 53 from Royden..taking math 435? (I went to UVic years ago).

f is continuous at a point if for every epsilon there is a delta..blah blah, you know the rest. For a hint, try to make the "for every" part handled by the intersection of your sets, that is take your sequence of sets to characterize the points were f satisfies a weaker condition than full blown continuity, but one that is getting "closer" to the full definition as you move along your sequence of sets.

3. Oct 7, 2004

### NateTG

Well, in topology you're going to want to take unions, not intersections of open sets. That said, I think you'll be better off if you work with neighborhoods rather than abstract topology definitions.

4. Oct 7, 2004

### Hurkyl

Staff Emeritus
Hrm, I spent a couple minutes thinking about the problem, and came up with the same approach as shmoe, so it must be right.

Nate, since he's after an element of $G_{\delta}$, and not an open set, it's okay to intersect his open sets. In fact, the theorem couldn't be true using just open sets, because there are functions continuous at exactly one point!

5. Oct 8, 2004

### fourier jr

yup... can't wait to get to the riesz-fischer theorem & related stuff (because i'm fourier jr)

thanks for the tips everybody; i think it worked out ok.