In a metric space , let and be disjoint closed subsets of . Prove that disjoint open sets and of such that and .

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- Aug 14th 2009, 05:43 PMKat-Mmetric space
In a metric space , let and be disjoint closed subsets of . Prove that disjoint open sets and of such that and .

- Aug 14th 2009, 06:15 PMputnam120
Let , then since they are disjoint and closed (I'll leave it to you to show that this is true). Then for each point create the neighborhood (neighborhood about x with radius r/3), similarly create for .

Finally take and . - Aug 14th 2009, 06:29 PMJose27
This is not true: Take the graph of in it is like two hyperbolas that get arbitrarily close as you approach (they both go to infinity). It is clearly closed (each 'hyperbola' is also closed), and the two 'hyperbolas' are disjoint but . for this to work either or must be compact

- Aug 14th 2009, 08:47 PMputnam120
I don't think that is closed. There is no point so that . Or is there some other definition of closed that you are using?

- Aug 14th 2009, 09:00 PMJose27
- Aug 15th 2009, 01:03 AMOpalg
- Aug 15th 2009, 12:24 PMKat-Mclosed set
- Aug 15th 2009, 12:41 PMOpalg
That is correct: if then . But putnam120's argument doesn't work for the reason pointed out by Jose27. The distance from B to A can be zero, as shown by the example of the two components of the graph of y=1/x^2. But the distance from any individual point in B to A must be strictly positive.

- Aug 15th 2009, 01:39 PMputnam120
My argument doesn't work because I took the over all pairs and . But

**Opalg**'s argument you look at a particular . To see that if just use the limit point definition of a closed set.

I have a question for**Jose27**: How are you able to see that my original argument would work if one of A or B were compact? - Aug 15th 2009, 03:53 PMEnrique2
Let me get into this nice discussion. If A is compact , is a continuous function, hence the infimum is attained (it is a minimum I mean). It can't be 0, if this were the case x would belong to since is assumed to be closed, contradicting the assumed disjointness.