Gδ set in normal space problem

[radou]

Homework Statement

Theorem. Let X be normal. There exists a continuous function f : X --> [0, 1] such that f(x) = 0, for x in A, and f(x) > 0 for x not in A, if and only if A is a closed Gδ set in X.

(A set in X is a Gδ set if it equals the intersection of a countable collection of open sets in X.)

The Attempt at a Solution

The proof is not complete yet, and I will point out the places where something is missing. I'd just like to verify if my current work is OK so I can fill in the gaps. Possibly without hints, since I still feel I can solve this one on my own.

==>

Let's first show that A is closed, under the hypothesis of the theorem. Let x be a point of X\A. Then f(x) > 0. Take a neighborhood V of f(x) contained in [0, 1], which doesn't contain 0. Since f is continuous, there exists a neighborhood U of x such that f(U) is contained in V. Since f(U) doesn't contain 0, we conclude that U and A are disjoint. Now do this for every x in X\A, and it follows that X\A equals a union of such neighborhoods. Hence, X\A is open.

Since A is closed and X is normal, and since X is an open neighborhood of A, we can find an open neighborhood U1 of A such that Cl(U1) is contained in X. Further on, for U1 and A, we can find an open neighborhood U2 of A such that Cl(U2) is contained in U1, etc. Now, from this we can conclude that A is contained in the countable intersection of open neighborhoods U1, U2, etc.

Now, I need to show somehow that A equals this intersection (if possible). I tried to assume there exists some element in the intersection and not in A, and arrive at a contradiction, but I still have nothing.

<==

Now, 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.

Let Sm be the set of all points of Am\A such that fm(x) = 0. My idea (possibly ridiculous) is to prove that there exists some positive integer k such that Sk is empty. Assume there doesn't. Then A alone contains a set of points of Am\A, which is a contradiction.

The last paragraph is something I really don't think will work, but still.

Thanks in advance.

Edit: actually, after thinking about it just a bit more, I concluded that the last paragraph is complete nonsense, hence it is to be ingnored. :) It would work only if such points would be in common for every single set of the intersection, and in general, they are different points.

 

[micromass]

I'll try not to give any hints (:

==>

Let's first show that A is closed, under the hypothesis of the theorem. Let x be a point of X\A. Then f(x) > 0. Take a neighborhood V of f(x) contained in [0, 1], which doesn't contain 0. Since f is continuous, there exists a neighborhood U of x such that f(U) is contained in V. Since f(U) doesn't contain 0, we conclude that U and A are disjoint. Now do this for every x in X\A, and it follows that X\A equals a union of such neighborhoods. Hence, X\A is open.

This is correct. However, I feel obliged to tell you that there is an easier proof. It is so that $$A=f^{-1}(\{0\})$$. And the inverse of a closed set is closed.

Since A is closed and X is normal, and since X is an open neighborhood of A, we can find an open neighborhood U1 of A such that Cl(U1) is contained in X. Further on, for U1 and A, we can find an open neighborhood U2 of A such that Cl(U2) is contained in U1, etc. Now, from this we can conclude that A is contained in the countable intersection of open neighborhoods U1, U2, etc.

Now, I need to show somehow that A equals this intersection (if possible). I tried to assume there exists some element in the intersection and not in A, and arrive at a contradiction, but I still have nothing.

Although this does not contain any mistakes, you will not get very far with it. I mean: your construction will not guarantee that A equals the intersection. Indeed, it is possible that I pick Un=X every time. Then the intersection of the Un is just X, which is clearly not what we want...

<==

Now, 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.

This is correct. You're on the right track here.

Let Sm be the set of all points of Am\A such that fm(x) = 0. My idea (possibly ridiculous) is to prove that there exists some positive integer k such that Sk is empty. Assume there doesn't. Then A alone contains a set of points of Am\A, which is a contradiction.

Paragraph ignored

 

[radou]

I'll try not to give any hints (:



This is correct. However, I feel obliged to tell you that there is an easier proof. It is so that $$A=f^{-1}(\{0\})$$. And the inverse of a closed set is closed.

It's fascinating how I manage to miss such obvious things.

Although this does not contain any mistakes, you will not get very far with it. I mean: your construction will not guarantee that A equals the intersection. Indeed, it is possible that I pick Un=X every time. Then the intersection of the Un is just X, which is clearly not what we want...

Good point.

OK, I'll do some more thinking.

 

[radou]

By the way, for the first part of the proof, perhaps it would be a better idea (since the normality of X implies that X is regular), to take A (which if proven closed) and any point in X\A, and find disjoint open neighborhoods of A and this point, and now use normality the same way I used it before? This guarantees that none of the neighborhoods of A will be X itself, but does it guarantee anything else?

 

[micromass]

Well, then it is possible that you end up with uncountably many sets. And you want a countable intersection...

 

[radou]

Hm, I see, but in general, there's something I don't understand here - start with a neighborhood U1 of A. Use normality to find a neighborhood U2 of A such that Cl(U2) is contained in U1. Can't I just "choose" (by this procedure, defined inductively) a countable number of such open neighborhoods Ui of A? And then A is contained in their countable intersection?

 

[micromass]

Yes, if you follow through the procedure you described, you will eventually end up with a countable collection. But the problem with that very procedure is that the intersection will not necessairily equal A.

This is not really a hint, but it might help to get some ideas straight: take a look at exercise 6 in the same chapter.

 

[radou]

Yes, if you follow through the procedure you described, you will eventually end up with a countable collection. But the problem with that very procedure is that the intersection will not necessairily equal A.

OK, I see. As a sub-question, (it may seem a bit of a stupid question, but I'd like to get this as straight as possible), this procedure will not eventually produce an intersection which equals A because there may exist clopen sets, right? i.e. the collection of neighborhood isn't really "contracting" aroung A in the same manner as, for example, the collection <-1/n, 1/n> equals {0}, in the countable intersection? I realize it is generally stupid to try to "visualize" such things, but still, I'm asking.

This is not really a hint, but it might help to get some ideas straight: take a look at exercise 6 in the same chapter.

OK, I'll do so.

 

[micromass]

No, visualizing is probably one of the best things you can do in topology! If I can't visualize a problem, then it suddenly becomes very hard for me...

As for your question. Yes, the existence of clopen sets will prevent you from picking the right intersection. But there are other things that might prevent you from doing so: you might be picking the sets to big! For example, let A=[0,1] in R. Then your procedure may yield the following sets $$]-1-1/n,2+1/n[$$. This will also be a bad choice of sets, since the intersection does not equal A. The reason why this will not work is exactly because the sets you picked were to big...

 

[radou]

OK, thanks, I feel I have more insight to the matter now.

 

[radou]

Hm, a relatively simple idea just occurred to me, could a right choice of sets be the family f^-1[0, 1/n>? A is definitely contained in their countable intersection. But vice versa?

A minute ago, I rushed about with a conclusion that, since for every n, [0, 1/(n+1)] is a proper subset of [0, 1/n>, their preimages will be proper subset also, but this would only hold if f had some other properties other than continuity, right? Such as injectivity, for example.

 

[radou]

Actually, let x be an element of $$\cap$$f^-1[0, 1/n>, and assume x is not in A. Then f(x) > 0, and for every n, f(x) < 1/n, which is impossible, right? Hence f(x) = 0 and x must be in A!

 

[micromass]

Hm, a relatively simple idea just occurred to me, could a right choice of sets be the family f^-1[0, 1/n>? A is definitely contained in their countable intersection. But vice versa?

Yes, you're certainly on the right track here. You'll have to use the function somehow. Try to prove the other unclusion.

What you were trying to show before you came up with this idea, is that a closed set in a normal space is Gdelta. But this is simply not the case. So you had to use the function f somewhere. But it seems that you've found it now...

A minute ago, I rushed about with a conclusion that, since for every n, [0, 1/(n+1)] is a proper subset of [0, 1/n>, their preimages will be proper subset also, but this would only hold if f had some other properties other than continuity, right? Such as injectivity, for example.

Hmm, I was thinking that surjectivity should be used that the preimages are proper subsets. I'm not sure were injectivity comes in...

 

[micromass]

Actually, let x be an element of $$\cap$$f^-1[0, 1/n>, and assume x is not in A. Then f(x) > 0, and for every n, f(x) < 1/n, which is impossible, right? Hence f(x) = 0 and x must be in A!

Yes, this is correct!

 

[radou]

Just a sub-question:

Let A1, A2, ... be a countable family of sets. If their intersection is empty, does it hold that there exists at least one pair of sets which are disjoint?

 

[micromass]

Not necessairily. Let $$X=\mathbb{Q}$$, and consider the family

$$A_n=]\pi-1/n,\pi+1/n[\cap \mathbb{Q}$$

then the intersection is empty, but no two sets are disjoint.

 

[radou]

OK, thanks.

I'm still trying to prove the other direction starting with what I wrote in post #1, but I find it a tough nut to crack for some reason. I'll let you know when I give up :D

 

[micromass]

Well, to be honest, the other implication is quite a hard one. I only know the answer because it's a little "trick" I've seen before. I'd be very impressed if you can find it without hints...

What you wrote in your OP is correct and is a step in a right direction. Somehow you have to combine all these functions in a way. But I'm not going to say more since you won't let me

 

[radou]

Somehow you have to combine all these functions in a way. But I'm not going to say more since you won't let me

Yes, I'm aware of that, and I'm working on it.

 

[radou]

Another sub-question:

If we have a sequence of continuous functions, can we define a new function as an infinite sum of these functions? Of course, this series should be convergent, for any x, right? Further on, does this series define a new continuous function?

(Again, no hints, I'm only poking around in all possible directions)

 

[micromass]

Ah, nice You're really getting there.
Indeed, if you have continuous functions $$f_n$$, then you can construct a new function

$$f(x)=\sum_{n=1}^{+\infty}{f_n(x)}$$

But, as you mentioned, there are two big problems
1) Our series may not converge. This is indeed possible: let $$f_n(x)=1$$ for a certain x and all n.
2) The sum may not be continuous. The sum of a series is just the limit of a sum. Of course, finite sums of continuous functions are always continuous. But a limit of continuous functions may not be continuous...

 

[radou]

The idea with such a series occurred to me before, but I didn't bother to do any investigation on it. It seems like an intuitive way to "elliminate" all the zero-points, since one of my other results (which I'll use if I manage to prove this completely) says that there is no point in x such that fm(x) = 0, for every positive integer m.

I managed to construct a function $$h(x)=\sum_{m=1}^\infty \frac{1}{2^m}f_{m}(x)$$, such that h(x) = 0 for x in A, and h(x) > 0 for x not in A, but as you mentioned, h isn't necessarily continuous! Unless I'm mistaken, this series is convergent (follows from the fact that the powers of 2 are convergent)?

But continuity remains a problem. I found out that this sequence of partial sums of this series should converge uniformly, in order for the series converge to a continuous function... I'll do some further research.

 

[radou]

Edit to the first paragraph: there is no point in X\A such that fm(x) = 0, for every positive integer m.

 

[micromass]

The series

$$h(x)=\sum_{n=1}^{+\infty}{\frac{1}{2^n}f_n(x)$$

is a very nice idea. It certainly deals with convergence issues...

There are certain theorems that give you conditions on when a series a continuous functions are continuous. I'm not going to tell them, since you don't want me to. But here's a small hint (I'll spoiler it, so you can chose to look at it. But it's a very silly hint)

Spoiler

All the theorems about continuousness are present in Munkres. But they're a bit obfusciated (I'm no native english speaker, so I'm not sure if that's even a word)...

 

[radou]

OK, thanks, I refuse to look at the hint yet. :)

I'm already strolling through the theorems... will post later, if I manage to conclude something.

 

[micromass]

Well, it's a very stupid hint... I guess you already came up with it yourself

 

[radou]

I tried a few uniform convergence tests, but they don't seem to be useful in this situation, so I'll try to do this in a primitive manner.

So, the idea is:

A series of functions converges uniformly is the sequence of partial sums of the series converges uniformly. If this happens to be true, and if all of our functions are continuous, then the sum of the series is a continuous function.

The sequence of partial sums Sn(x) converges uniformly to the sum S(x) if for every ε > 0 there is a positive integer N (dependent only on ε), such that for all n >= N and for all x in X, |Sn(x) - S(x)| < ε holds.

In our case, $$S_{n}(x) = \sum_{k=1}^n \frac{1}{2^k} f_{k}(x)$$ and $$S(x) = \sum_{n=1}^{\infty} \frac{1}{2^k} f_{k}(x)$$.

Let ε > 0 be given. Then |Sn(x) - S(x)| = $$\left| \sum_{k=n+1}^{\infty}\frac{1}{2^k} f_{k}(x) \right|$$.

Now, this is the part I'm not really sure about, because I tried to do it intuitively and fiddling around with series in Mathematica: the greatest possible value for any function fk is 1, and the powers of 2, in this situation, converge to 0, and hence, whatever the value of fk is and for any k, we can find an integer large enough so that the term |S(x) - Sn(x)| < ε.

If this is true, then the series converges uniformly and its sum is a continuous function, what we are looking for.

 

[micromass]

May I ask you what uniform convergence tests you tried out?

Your reasoning is very sound. But there is a uniform convergence test that generalizes what you are trying to do.

 

[radou]

Of course.

The Weierstrass M-test required a boundedness criteria which didn't seem to match, and the Abel test had a decreasing criteria which didn't seem to work...Anyway, it's very likely that I missed something, since it's 0:23 am here and I'm getting a bit tired :)

 

[micromass]

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

 

[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. :)