Strong form of the Urysohn lemma

[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!