Strong form of the Urysohn lemma

[radou]

Homework Statement

Let X be a normal space. There exists a continuous function h : X --> [0, 1] such that h(A) = 0, h(B) = 1 and h(X\(AUB))$$\in$$<0, 1> iff A and B are disjoint closed Gδ sets in X.

The Attempt at a Solution

I'll only display my attempt for one direction, since the other one is rather easy.

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Let A and B be disjoint closed Gδ sets in X. Since their union is a Gδ set, too, there exists a function f : X --> [0, 1] such that f(AUB) = 0 and f(X\(AUB)) > 0. Next, for every positive integer m, apply the Urysohn lemma to the sets X\Bm (since B is Gδ, the sets Bm are open members of the family whose countable intersection equals B) and B in order to obtain a continuous function gm : X --> [0, 1] such that gm(X\Bm) = 0 and gm(B) = 1.

Now, for every positive integer, define the function hm(x) = 1/2 f(x) + gm(x). I'm not really sure about this, but the function we're looking for could be h(x) = lim hm(x), as n --> ∞. If x is in A, for m large enough (since Bm could actually intersect A) hm(A) = 0, hm(B) = 1, and any x not in A or B should eventually converge to 1/2f(x), which is in <0, 1>.

I hope this works, thanks in advance.

 

[micromass]

Hmm, I don't know if that would work, but I do have some small issues with the proof.

I'm not really sure about this, but the function we're looking for could be h(x) = lim hm(x), as n --> ∞.

It looks like a nice choice, but why is h continuous?

If x is in A, for m large enough (since Bm could actually intersect A) hm(A) = 0

Even though A and B are disjoint, it is possible that every A and Bm are not disjoint. This is possible since I can pick the Bm in a really evil way.

For example: A=[0,1], B=[2,3]. Then I can pick $$B_0=]1/2,4[$$ and $$B_n=]2-1/n,3+1/n[\cup ]0,1/4[$$.

 

[radou]

Hmm, I don't know if that would work, but I do have some small issues with the proof.



It looks like a nice choice, but why is h continuous?

Hm I thought it was, since hm(x) converges either to the sum of f and g, which is continuous, or to f, which is continuous. Depends on the choice of our x. I'll have to study this in more detail, but I like the idea very much.

Even though A and B are disjoint, it is possible that every A and Bm are not disjoint. This is possible since I can pick the Bm in a really evil way.

For example: A=[0,1], B=[2,3]. Then I can pick $$B_0=]1/2,4[$$ and $$B_n=]2-1/n,3+1/n[\cup ]0,1/4[$$.

Yes, that's a good point. It seems I'll ahve to use normality here.

I'll update after giving it a second thought.

 

[radou]

Well, the issue about the disjointness of A and the Bm's could be resolved in this way (I hope):

Since X is normal, choose disjoint neighborhoods U and V of A and B, respectively.

Claim. There exists some positive integer m such that B$$\subseteq$$Bm$$\subseteq$$V.

Assume the contrary. Then, for every m, V$$\subseteq$$Bm and hence V would be contained in the intersections of the Bm's, which is impossible.

 

[micromass]

Well, the issue about the disjointness of A and the Bm's could be resolved in this way (I hope):

Since X is normal, choose disjoint neighborhoods U and V of A and B, respectively.

Claim. There exists some positive integer m such that B$$\subseteq$$Bm$$\subseteq$$V.

Assume the contrary. Then, for every m, V$$\subseteq$$Bm and hence V would be contained in the intersections of the Bm's, which is impossible.

No, I fear that this won't work. My example above shows that this is not always true.
What's wrong with your proof is that the contrary of $$B_m\subseteq V$$ is not $$V\subseteq B_m$$...

What you have to do, is choose a special kind of $$B_m$$...

 

[radou]

That's what I feared.

Btw, what would be the contrary of my claim? Is it possible to formulate the contrary? When I think about it, the contrary would be "there does not exist a positive integer m such that Bm$$\subseteq$$V holds", which means that for every m, V is contained in Bm? What's wrong with the reasoning here, could you clarify?

 

[radou]

Regarding my defined sequence of functions {hm(x)}, I should try to see if it converges uniformly, since then h would be continuous.

 

[micromass]

That's what I feared.

Btw, what would be the contrary of my claim? Is it possible to formulate the contrary? When I think about it, the contrary would be "there does not exist a positive integer m such that Bm$$\subseteq$$V holds",

This is correct. Sadly, there is not really much more you can say then this.
Although, sometimes this is useful: the contrary of "there exists an m such that $$B_m\subseteq V$$" is "forall m, it holds that $$B_m\setminus V$$ is not empty".

which means that for every m, V is contained in Bm? What's wrong with the reasoning here, could you clarify?

Consider B=[0,2] and V=[1,3]. Then it is not true that $$B\subseteq V$$. You claim that the contrary of this is $$V\subseteq B$$, but this is not so, since in this case it is not true that $$V\subseteq B$$.

Your confusion probably comes from the total ordering relation. In $$\mathbb{R}$$, it is indeed true that the contrary of x<y is $$y\leq x$$. But this is only so, because the order is total.
The order $$\subseteq$$ is not a total order, thus the above does not hold.

 

[micromass]

Regarding my defined sequence of functions {hm(x)}, I should try to see if it converges uniformly, since then h would be continuous.

To be honest, I really doubt that the convergence is uniform. You know almost nothing about the function $$g_m$$. It's not even clear if this series converges pointswise!

I'd suggest changing your sequence in such a way that uniform convergence becomes obvious...

Note: I don't immediately see what the proof is myself, so take my advice with a grain of salt. I could very easily be wrong...

 

[radou]

This is correct. Sadly, there is not really much more you can say then this.
Although, sometimes this is useful: the contrary of "there exists an m such that $$B_m\subseteq V$$" is "forall m, it holds that $$B_m\setminus V$$ is not empty".

Hm, but doesn't this (i.e. the contrary) imply that for every m, the intersection of Bm and V is non-empty, and hence the intersection of all the Bm's and V is non-empty too? (this is probably wrong, but I'm curious).

 

[micromass]

No, take Bm=[0,2] and V=[4,6], then the intersection is empty, while Bm/V=[0,2] is nonempty...

 

[radou]

No, take Bm=[0,2] and V=[4,6], then the intersection is empty, while Bm/V=[0,2] is nonempty...

OK...these examples from the reals are very useful when checking such things, I should think more like this...

 

[radou]

Even though A and B are disjoint, it is possible that every A and Bm are not disjoint. This is possible since I can pick the Bm in a really evil way.

For example: A=[0,1], B=[2,3]. Then I can pick $$B_0=]1/2,4[$$ and $$B_n=]2-1/n,3+1/n[\cup ]0,1/4[$$.

OK, I know I'm becoming very boring now, but I still don't get something in this example:

In your case, the intersection of the Bn's would be <0, 1/4>U[2, 3]. But this does not equal B, right? And we know that B is G-delta, and for the Bn's we choose exactly the sets whose intersection equals B, not?

 

[radou]

Ahh, pardon me, ignore this! I forgot about B0! :)

 

[micromass]

But the intersection of the Bn (with n including 0) does equal B. If you exclude 0, then you are correct that the intersection is not B...

 

[micromass]

But you can easily modify the Bn such that they are all disjoint from A, and such that the intersection does equal B...

 

[radou]

But the intersection of the Bn (with n including 0) does equal B. If you exclude 0, then you are correct that the intersection is not B...

Yes, yes, I know, my apologies, I was being hasty again to conclude something stupid. I'm going to try to reformulate this whole idea now.

 

[radou]

But you can easily modify the Bn such that they are all disjoint from A, and such that the intersection does equal B...

Well, we could choose disjoint sets U and V containing A and B, and create a new family of open sets whose intersection equals B, i.e. for every m, B'm = V$$\cap$$Bm, right?

 

[micromass]

Well, we could choose disjoint sets U and V containing A and B, and create a new family of open sets whose intersection equals B, i.e. for every m, B'm = V$$\cap$$Bm, right?

Yes that is correct. But note that you do not even need normality for this. You could also do B'm=Bm/A. But now I'm just being pedantic

So, the only problem remaining is the continuity of h...

 

[radou]

Yes that is correct. But note that you do not even need normality for this. You could also do B'm=Bm/A. But now I'm just being pedantic

So, the only problem remaining is the continuity of h...

Yes, I recall from somewhere in the book the fact that the difference of an open and a closed set is open again...

 

[radou]

OK, now that we reformulated the sets Bm to Bm', we could define:

$$h(x) = \sum_{m=1}^{\infty}\frac{1}{2^m}\left(\frac{1}{2}f(x)+g_{m}(x)\right)$$

If x is in A, h(x) = 0, if x is in B, h(x) = 1, but the problem is if x is in Bm\B (from now on the Bm's are referred to as Bm, for convenience). I should somehow limit the value of h(x) to something less than 1 at these points, and I may have a few ideas.

Does it make sense to ask oneself if gm(x) < 1 for all positive integers m greater than some integer m0?

 

[micromass]

Does it make sense to ask oneself if gm(x) < 1 for all positive integers m greater than some integer m0?

Well, if x is in Bm\B, then it can even happen that gm(x)=1 for every m. So I'm not sure if this is the right way to approaching the problem.

 

[radou]

Well, if x is in Bm\B, then it can even happen that gm(x)=1 for every m. So I'm not sure if this is the right way to approaching the problem.

Yeah, that's what I feared...I'll have to find a new approach, won't get away with the one from the last theorem :)

 

[micromass]

OK, I think I have finally found a proof for this exercise. But it's something completely else then what you're trying to do (not that I'm saying that your method won't work).

My proof doesn't use exercise 4, but it uses the following stronger version of exercise 4:
Let X be normal. Let A be a closed $$G_\delta$$ set and B closed. Then there exist a continuous function f:X-->[0,1] such that f(x)=0 if x in A, f(x)>0 if x is not in A, f(x)=1 if x is in B.

 

[radou]

Btw, just to clear something up, is the "old function", defined with h(x) = lim (1/2 f(x) + gm(x)) (as n --> infinity), well-defined at all? Does gm --> 0, for some x outside B?

 

[radou]

It seems not, since you said gm(x) can equal 1 for any integer m, so by definition, the limit can't be 0?

 

[micromass]

Btw, just to clear something up, is the "old function", defined with h(x) = lim (1/2 f(x) + gm(x)) (as n --> infinity), well-defined at all? Does gm --> 0, for some x outside B?

No, it is indeed not clear if the limit even exists! It is true that h is well defined on A and B, but the limit doesn't necessarily exist outside these sets.

There is another problem. If h does exist and is continuous, then it is not guaranteed that h(x)<=1...

 

[radou]

I'm trying to prove the "stronger" version of ex. 4 you used, but I don't see right away how to prove the theorem even with this fact...but no hints yet, please :)

this one seems a bit harder than ex. 4 (I mean the strong version of U.l.)...

 

[radou]

I sense somehow that the pasting lemma could be used here in a way...btw, it works only if both sets on which the functions are defined are both closed or both open, right? (and the union of these sets must equal X, and they must have a non-empty intersection, i.e. for any x in the intersection f(x) = g(x) must hold)

 

[micromass]

I don't know if the pasting lemma is necessary, but it sure could come in handy.
But note that the pasting lemma also hold if the intersection is empty. Thus the only conditions are that the sets are both closed/open and that the union is X.

 

[radou]

I don't know if the pasting lemma is necessary, but it sure could come in handy.
But note that the pasting lemma also hold if the intersection is empty. Thus the only conditions are that the sets are both closed/open and that the union is X.

Hm, that's a useful piece of information..in Munkres, page 108. the intersection seems to be a crucial part of the formulation of Theorem 18.3., so I thought it won't work with it being empty .

 

[radou]

But, of course, if their intersection happens to be non-empty, f(x) = g(x) must hold, for every x in the intersection, right?

 

[micromass]

Yes, but Munkres does not require the intersection to be nonempty there. He simply states: IF the intersection is nonempty, then f(x)=g(x) for every x in the intersection. But if the intersection is empty, then the theorem holds without requirement.

 

[micromass]

But, of course, if their intersection happens to be non-empty, f(x) = g(x) must hold, for every x in the intersection, right?

Correct, that's always necessary!

 

[radou]

OK, I think I have finally found a proof for this exercise. But it's something completely else then what you're trying to do (not that I'm saying that your method won't work).

My proof doesn't use exercise 4, but it uses the following stronger version of exercise 4:
Let X be normal. Let A be a closed $$G_\delta$$ set and B closed. Then there exist a continuous function f:X-->[0,1] such that f(x)=0 if x in A, f(x)>0 if x is not in A, f(x)=1 if x is in B.

It seems to me that this "stronger version" of exercise 4 can be proved basically the same way as ex. 4., with the exception that the Am's are disjoint from B. Since from the definition of the series of function in the proof of ex. 4. it follows that f(B) = 1, for any x in B, not? And f(A) = 0 for x in A, and f(x) > 0, for x not in A.