Gδ set in normal space problem

[radou]

Umm, the Weierstrass-M-test didn't work?? You might want to rethink that

Well, perhaps my resources were a bit bad.

I hate to mention it, but I looked on wikipedia, too :)

http://en.wikipedia.org/wiki/Weierstrass_M-test

according to this, my constant Mn = 1, for every n and x, right? And the series of "1's" does not converge.

Now that I look at Kurepa's analysis book, I find a more sophisticated formulation. Briefly, it says that if every member fk of a series of functions is less (by absolute value) than the corresponding member of a convergent series of positive numbers, then the series of functions fk converges uniformly.

Actually, this is something I need, since I introduced the power series..Hm.

 

[micromass]

Yes, but in this case you have to find a M_n such that |2^{-n}f_n(x)|\leq M_n. Your series is

\sum{2^{-n}f_n(x)}

isn't it? If the series were

\sum{f_n(x)}

then I'd agree with you. But you've added the 2^{-n}...

 

[radou]

Yes, but in this case you have to find a M_n such that |2^{-n}f_n(x)|\leq M_n. Your series is

\sum{2^{-n}f_n(x)}

isn't it? If the series were

\sum{f_n(x)}

then I'd agree with you. But you've added the 2^{-n}...

This is exactly what I was talking about - lack of concentration :)

In this case, it's simple - we end up with a series 1/2, 1/4, etc. i.e. the series of powers of 2, which converges, right?

 

[micromass]

Yes, that is correct! You've actually solved this problem, I'm amazed I actually teach topology to a class of undergraduates and I don't think any of them would have found this proof. I think you're a very good mathematician



That said, I want to make some remarks:

- another way to make one function out of the fⁿ, is by taking the supremum. Thus f(x)=\sup_{n\in \mathbb{N}}{f_n(x)}. But showing continuity is (imo) a a little more tedious then with the series.

- a foreshadowing for later: you will use the same technique in the proof of the Tietze extension theorem. But there, the construction of the right series will be a bit more difficult.

 

[radou]

wow, thanks a lot!

I'll post the whole solution tomorrow. Right now, I can't think of anything anymore, but I'll look at the alternative proof tomorrow, too. :)

 

[micromass]

Hmm, forget about my alternative "proof". It obviously doesn't work. I guess it's getting late for me to

 

[radou]

Here, for convenience, I'll just sum up the sketch of the proof of the other direction.

<==

Let A be a closed Gδ set in X, so it equals a countable intersection of open sets in X. For any positive integer m, define Am = \cap_{i=1}^m A_{i}. Further on, for any positive integer m, apply the Urysohn lemma to the disjoint closed sets A and X\Am in order to obtain a continuous function fm such that fm(A) = 0 and fm(X\Am) = 1. So, we obtain a sequence of continuous functions {fm}.

Now, for every positive integer m, define Sm = {x in Am\A : fm(x) = 0}. Then fm^-1({0}) = A U Sm\subseteqAm, for every m. Now, \cap_{m}(A U Sm) = A U (\cap_{m}Sm) = A, which implies \cap_{m}Sm = \emptyset. So, for any x in X, there exists some positive integer m such that fm(x)\neq0. This motivates the following definition:

Define h(x) = \sum_{k=1}^{\infty}\frac{1}{2^k}f_{k}(x). It is easily checked that for any x in A, h(x) = 0, and for any x in X\A, h(x) > 0. Also, one can show that this series converges uniformly, and that its sum (i.e. h(x)) is a continuous function.

 

[micromass]

Looks good!

 

[radou]

OK, thanks again for a great guideance!