Strong form of the Urysohn lemma

[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.

 

[micromass]

Yes, you are correct. The stronger version basically has thesame proof. There are just some minor tweaks...

 

[radou]

Just a thought, is the characteristic function for a subset B of a topological space X continuous on X?

 

[micromass]

No, not in general. We even have the following equivalence:

"the characteristic function of B is continuous if and only if B is open and closed"

 

[radou]

So, we could formulate this theorem:

Let X be normal, with A Gδ and B clopen. Then there exists a function f : X --> [0, 1] such that f(A) = 0, f(B) = 1 and 0 < f(x) < 1, for x in X\(AUB).

The proof follows by the "strong" version of ex.4., and if we define k(x) to be the characteristic function of the set B, then h(x) = 1/2 f(x) + 1/2 k(x) is the function we're looking for.

 

[micromass]

Yes, this would be correct

h(x) = 1/2 f(x) + 1/2 k(x) is the function we're looking for.

Now, in the general case (thus if B is not necessarily open), you could essentially do the same thing. Except that you can not choose k the characteristic function. But maybe you can also derive a k from the "strong" version of ex. 4...

 

[radou]

I'm thinking about this problem again. No hints yet, but just out of curiosity, is there a "trick" involved in this one too?

 

[micromass]

Uh, depends on what you call a trick. I think there might be a small trick involved...

You will actually have to use the "strong" version of exercise 4 twice:
1) you find a function which is 0 precisely on A, and which is 1 on B.
2) you find a function which is 0 precisely on B, and which is 1 on A.

Now you have to combine these functions. That might involve a small trick, but I think you can find it on yourself...

 

[radou]

I smell Theorem 18.2.(f) (page 108 in Munkres) could resolve this situation somehow..

 

[radou]

Here, finally:

if we use the "stronger" form of ex. 4, we have function f and g satisfying:

f(A) = 0, f(B) = 1, g(A) = 1, g(B) = 0.

Then h(x) = f(x)/(1 + g(x)) is the desired function!

 

[micromass]

Ah yes, you are correct!