Completely regular space problem

[radou]

Homework Statement

For some reason this one is really troubling me. So, a small hint (if possible) would be highly appreciated.

Let X be completely regular, and let A and B be disjoint closed subsets of X. If A is compact, there exists a continuous function f : X --> [0, 1] such that f(A) = {0} and f(B) = {1}.

The Attempt at a Solution

I've tried to solve this in different ways, but I don't seem to be getting anywhere. I don't see how compactness helps here - there's either something really obvious I'm missing here or a dubious result from one of the previous exercises which should be used here.

 

[ystael]

In a completely regular space, a point can be separated from a closed set by a continuous function. The compactness of $$A$$ tells you that $$A$$ is "almost" a point, i.e., can be approximated by pasting together things that happen at finitely many points.

 

[micromass]

Well, the easiest proof of this theorem is by considering the Cech-Stone compactification, but I guess you don't have that tool available yet.

I'm giving you two hints. I hope that it'll help you without giving away to much

HINT 1:

Spoiler

Show that it is enough to find a function f:X-->[0,1] such that f(B)=1 and f(A) a subset of [0,1/3].

HINT 2:

Spoiler

For every x in A, we can find a function fx:X-->[0,1] such that f(x)=0 and f(B)=1 (by complete regularity). Then we have {$$f_x^{-1}$$([0,1/3[) | x in X} which is an open cover of A.

 

[radou]

Yes, I'm aware that this "finiteness" will make it possible to construct a continuous function out of finitely many continuous ones, perhaps.

My first aim is:

Since X is completely regular, for B and for any x in A there exists a continuous function f : X --> [0, 1] such that f(B) = {0} and f(x) = 1. Then g(x) = 1 - f(x) is continuous and g(B) = 1, along with g(x) = 0.

Since A is compact, let U1, ..., Un be a finite cover for A. For every Ui, pick an xi contained in it. Construct the function gi : X --> [0, 1] like in the upper paragraph such that gi(B) = 1 and gi(xi) = 0.

That's an idea, but I don't see right away where it leads to.

 

[radou]

Ah, now I saw your reply, micromass. Will think about it, too.

 

[radou]

Btw, I didn't come across the Cech-Stone compactification yet, yes.

 

[radou]

Yes, the set fx1^-1([0, 1/3>), ... ,fxn^-1([0, 1/3>) is then a finite subcover for A. So, any y in A is contained in at least one of these open sets, say fxj^-1([0, 1/3>). In this set, the element xj maps to 0 and fxj(y) is contained in [0, 1/3>. But what else does this say? Gee, I feel slow today...

 

[micromass]

Well, you could modify your $$f_{x_j}$$ such that all elements in $$f_{x_j}^{-1}([0,1/3[)$$ map to 0.

Then you should be able to combine all your functions. But that's not really a problem, since you only have a finite amount of functions...

 

[radou]

Hmm, just a sec, something different just occurred to me. It probably won't work, but I'm interested in the mistake in reasoning.

Start with constructing the family {fx} of functions with the desired property. Then arrive at a finite cover for A with fx1^-1([0, 1/n>), ... ,fxm^-1([0, 1/n>). We can do this for every n = 1, 2, ... It follows that for any n, and any x in A, there exists a function fxj such that fxj(x) is contained in [0, 1/n>. But this doesn't mean that there exists a function from this countable family which maps x to 0, right?

Just out of curiosity, if such a function existed, could one define a countable product of all these functions which would be continuous? Since then, for any x in A, one of these functions would vanish at x. Or is it not possible to do so?

 

[micromass]

Hmm, just a sec, something different just occurred to me. It probably won't work, but I'm interested in the mistake in reasoning.

Start with constructing the family {fx} of functions with the desired property. Then arrive at a finite cover for A with fx1^-1([0, 1/n>), ... ,fxm^-1([0, 1/n>). We can do this for every n = 1, 2, ... It follows that for any n, and any x in A, there exists a function fxj such that fxj(x) is contained in [0, 1/n>. But this doesn't mean that there exists a function from this countable family which maps x to 0, right?

Yes, it doesn't mean that there is a function from these family such that x maps to 0. I can't give you a counterexample right away, but I'm certain that it doesn't hold.

Just out of curiosity, if such a function existed, could one define a countable product of all these functions which would be continuous? Since then, for any x in A, one of these functions would vanish at x. Or is it not possible to do so?

Well, even if such a function existed, then we could indeed define a countable product of these functions. But you're working with an infinite product, so there could potentially be convergence issues (outside A and B of course).

 

[radou]

Well, even if such a function existed, then we could indeed define a countable product of these functions. But you're working with an infinite product, so there could potentially be convergence issues (outside A and B of course).

OK, just another off-topic subquestion, theoretically, there could be a countable product of continuous functions which is continuous and well-defined (i.e. convergent), right?

 

[micromass]

Yes, there could be a countable product which is continuous and convergent. Such a functions are particularly usefull is complex analysis. But I don't think you'll ever meet such monsters in topology :tongue2:

 

[radou]

Here's an idea, but I'm afraid there will be convergence issues preventing the function to be continuous, unless I'm mistaken.

So, for any y in A there exists some function fxj which maps y into the interval [0, 1/3>. Define the product f = fx1*...*fxn, which is a continuous function. Now define g(x) = lim f(x)^n, as n --> ∞. Indeed, for x in A, f(x) is in [0, 1/3> (since at least one of the factors is in this interval), and so g converges to 0. If x is in B, g converges to 1. But I'm not sure if this convergence is uniform. Let ε > 0 be given. We wish to show that there exists some positive integer N such that for n >= N, |fn(x) - g(x)| < ε, for any x in X. If x is in A, such N exists. If x is in B, it's satisfied trivially. If x is in X\(AUB), and if fxi(x) < 1, for some i, f(x)^n converges to 0, and g(x) = 0, so such N exists. If fxi(x) = 1 for all i, f(x)^n converges to 1, to it seems to be okay again.

I either solved this, or there's some great misconcept from my side.

 

[ystael]

The example where $$X = [0, 1], A = \{0\}, B = \{1\}, f(x) = x$$ shows you that you cannot use this "limit of powers" construct to give a continuous function, since $$x \mapsto \lim_{n\to\infty} (f(x))^n$$ is the characteristic function of $$\{1\}$$. The argument you gave does not establish uniform convergence, only pointwise convergence.

You can do this with a rather cruder manipulation of the functions. As a small hint, remember that you only need your function to be continuous, not differentiable!

 

[radou]

The example where $$X = [0, 1], A = \{0\}, B = \{1\}, f(x) = x$$ shows you that you cannot use this "limit of powers" construct to give a continuous function, since $$x \mapsto \lim_{n\to\infty} (f(x))^n$$ is the characteristic function of $$\{1\}$$. The argument you gave does not establish uniform convergence, only pointwise convergence.

..and the characteristic function of a set is continuous iff the set is clopen, which is not our case, right?

You can do this with a rather cruder manipulation of the functions. As a small hint, remember that you only need your function to be continuous, not differentiable!

OK, I'll think about it.

 

[micromass]

But, can't you transform the fxj such that it is 0 on $$f^{-1}_{x_j}([0,1/3[)$$?
Maybe take the maximum of this function and something else...

 

[radou]

But, can't you transform the fxj such that it is 0 on $$f^{-1}_{x_j}([0,1/3[)$$?

Yes, I'm trying to do this...

Maybe take the maximum of this function and something else...

The maximum? "Mathematician's" block :uhh: Btw, you mean the supremum?

 

[micromass]

Yes, I mean the supremum. But in this case, it will be the maximum...

 

[radou]

Yes, I mean the supremum. But in this case, it will be the maximum...

You mean, since A is compact? And if we look at the restrictions of our functions to A?

 

[micromass]

No, I mean that you should define a new function like this:

$$h(x)=\max\{f(x),1/3\}$$

Try doing things like this...

 

[radou]

No, I mean that you should define a new function like this:

$$h(x)=\max\{f(x),1/3\}$$

Try doing things like this...

Hm, you mean something like this..

If x is in A, there exists some fxj which maps x into [0, 1/3>. It follows that for any x in A, hj(x) = max{fxj(x), 1/3} = 1/3. I'll try to develop this further, but is hj(x) defined this way (even if this is not the right way to define it, I'm still interested) continuous?

 

[radou]

Actually, hj(x) should be continuous, since:

i) fxj^-1([0, 1/3]) is closed, and for any x in this set, hj(x) = 1/3, which is continuous
ii) fxj^-1([1/3, 1]) is closed, and for any x in this set hj(x) = fj(x), which is continuous
iii) the intersection of these sets is fxj^-1({1/3}), ahd here hj(x) = 1/3 = fj(x)

Hence, by the pasting lemma, hj(x) is continuous, right?

 

[micromass]

Yes, you are completely correct! The function is continuous by the pasting lemma.

 

[radou]

OK, then:

for each hj(x), the composition of hj(x) and the function F(x) = 3/2 x - 1/2 i.e. (F o hj)(x) maps elements of A into 0 and elements of B into 1.

If this is correct, I only need to find a way to combine all these functions and that's it.

 

[radou]

Which is trivial! Consider the product of the functions (F o hj)(x). It is continuous, and if x is in A, for some j, the function vanishes, and so does the product. If x is in B, it equals 1. Right?

 

[micromass]

Yes, I think you've got it! Congratulations

I suggest you remember the trick about the maximum, it's useful a lot of times!

 

[radou]

OK, thanks!

To be honest, I doubt I'd come up with that "maximum" trick, but it seems very useful!

By the way, this exercise section is most definitely the hardest one yet. I don't know if it's only me, but there are just (if even) a few exercises here which can be solved in a "straightforward" manner...But I guess that's the point about mathematics.

 

[micromass]

Well, it's OK if you can't come up with the maximum trick. To be honest, I only know it because I've seen it somewhere. I guess I wouldn't have found it otherwise.

And you're 100% right that these exercises are quite hard. But, to be honest, I've always found things like complete regularity and Cech-Stone compactifications quite difficult. Maybe that is so since there is no way to visualize these things.

The thing with topology (and many other mathematic fields) is that there are a lot of tricks involved. Once you know most of the tricks, the exercises become easy. But if you don't know them, then exercises can be very very very difficult. But to learn about those tricks, you can only make exercises, I'm afraid... I think it is very good that you try to make all these exercises, once you've finished Munkres, then you've got a very solid foundation of topology and analysis...

 

[radou]

But to learn about those tricks, you can only make exercises, I'm afraid... I think it is very good that you try to make all these exercises, once you've finished Munkres, then you've got a very solid foundation of topology and analysis...

Yes, it's probably completely senseless to do mathematics without solving problems. I think I'd literally learn nothing.

Thanks again to you and ystael! 'till soon..