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 .
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
so if is an elemnt of , since . Is that correct? if so, doesnt putnam120's argument work?
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.
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?
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.