Analysis: continuous function and open sets

[malicx]

Homework Statement

Let (X, p) be a metric space, and let A and B be nonempty, closed, disjoint subsets of X.
define d(x,A) = inf{p(x, a)|a in A}

h(x) = d(x, A)/[d(x, A) + d(x, B)]
defines a continuous function h: X -> [0,1]. h(x) = 0 iff x is in A, and h(x) = 1 iff x is in B. Infer that there exist open sets U and V of X such that $$A \subset U$$ and $$B \subset$$V with $$U \cap V = \emptyset$$

Homework Equations

The Attempt at a Solution

I have showed everything except the last part about disjoint open sets. I don't think I can just say that since A and B are disjoint, there is an r>0 such that $$B_r(A) \cap B_r(B) = \emptyset$$. I'm actually having a hard time seeing how the iff statements are necessary. Any hints would be helpful.

 

[micromass]

For your general culture: the function h is called a Urysohn function.

Now for the proof. Consider $$h^{-1}([0,1/2[)$$ and $$h^{-1}(]1/2,1])$$. These are the sets you're looking for...

 

[malicx]

For your general culture: the function h is called a Urysohn function.

Now for the proof. Consider $$h^{-1}([0,1/2[)$$ and $$h^{-1}(]1/2,1])$$. These are the sets you're looking for...

I'm not sure I understand... we know that, given an open set in [0, 1], its inverse image is open in X by continuity. But [0, 1/2) and (1/2, 1] are neither open nor closed. If we say (0, 1/2) and (1/2, 1) then we are saying that h(x) =/= 0, so x is not in A, but we are looking for $$A \subset U$$ for some U, right?

 

[micromass]

Remember that your codomain is [0,1]. The sets [0,1/2[ and ]1/2,1] are open in [0,1] (they are not open in R of course, but they are in [0,1].

If you don't like that, then you can always consider the sets ]-1,1/2[ and ]1/2,2[ and take R as codomain of h...

 

[malicx]

Remember that your codomain is [0,1]. The sets [0,1/2[ and ]1/2,1] are open in [0,1] (they are not open in R of course, but they are in [0,1].

If you don't like that, then you can always consider the sets ]-1,1/2[ and ]1/2,2[ and take R as codomain of h...

Of course it is! >_<. I am really bad at topology...

Thank you for your help, I'm confident I understand it now (and it is so obvious!)