Analysis: continuous function and open sets

In summary, a Urysohn function h(x) = d(x, A)/[d(x, A) + d(x, B)] defines a continuous function h: X -> [0,1], where h(x) = 0 iff x is in A, and h(x) = 1 iff x is in B. This implies the existence of open sets U and V of X such that A \subset U and B \subset V with U \cap V = \emptyset. These sets can be found by considering the inverse images of [0,1/2[ and ]1/2,1] under h, which are open in [0,1] and thus open in X. Alternatively, the sets
  • #1
malicx
52
0

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 [tex]A \subset U[/tex] and [tex]B \subset [/tex]V with [tex]U \cap V = \emptyset [/tex]

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 [tex]B_r(A) \cap B_r(B) = \emptyset[/tex]. I'm actually having a hard time seeing how the iff statements are necessary. Any hints would be helpful.
 
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  • #2
For your general culture: the function h is called a Urysohn function.

Now for the proof. Consider [tex] h^{-1}([0,1/2[) [/tex] and [tex]h^{-1}(]1/2,1])[/tex]. These are the sets you're looking for...
 
  • #3
micromass said:
For your general culture: the function h is called a Urysohn function.

Now for the proof. Consider [tex] h^{-1}([0,1/2[) [/tex] and [tex]h^{-1}(]1/2,1])[/tex]. 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 [tex]A \subset U[/tex] for some U, right?
 
  • #4
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...
 
  • #5
micromass said:
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!)
 

1. What is a continuous function?

A continuous function is a type of mathematical function that describes a relationship between two variables in which small changes in one variable result in small changes in the other variable. It means that the graph of the function has no breaks or discontinuities.

2. How do you determine if a function is continuous?

A function is continuous if it meets the three criteria of continuity: it is defined at the point in question, the limit of the function at that point exists, and the limit is equal to the value of the function at that point. In other words, the function must have no breaks or holes in its graph.

3. What is an open set in mathematical analysis?

An open set is a subset of a metric space where every point in the set has a neighborhood that is also contained within the set. This means that there are no boundary points in the set, and all points within the set can be "approached" from all directions without leaving the set.

4. How do you prove that a set is open?

To prove that a set is open, you must show that for any point in the set, there exists a small enough open ball (or neighborhood) around that point that is also contained within the set. This can be done by using the definition of an open set and showing that every point in the set has a neighborhood that is also contained within the set.

5. What is the relationship between continuous functions and open sets?

In mathematical analysis, continuous functions are often defined as those functions whose preimage of any open set is also open. This means that the inverse image of an open set under a continuous function will always be an open set. Additionally, in topology, open sets are used to define the concept of a continuous function.

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