A and B disjoint, A compact, and B closed implies there is positive distance between both sets

Claim: Let X be a metric space. If A,BX are disjoint, A is compact, and B is closed, then there is δ>0 so that |αβ|δαA,βB.

Proof. Assume the contrary. Let αnA,βnB be chosen such that |αnβn|0 as n.

Since A is compact, there exists a convergent subsequence of αn(nN), αnm(mN), which converges to αA.

We have

|αβnm||ααnm|+|αnmβnm|0asm.
Hence α is a limit point of B and since B is closed αB, contradiction.

Is my proof correct? I feel as though I am missing something simple which would trivialize the proof.

Answer

For the sake of having an answer addressing your question:

Yes, your proof is perfectly okay and I don’t think you can get it any cheaper than you did it.


Let me expand a little on what you can do with these arguments (also providing details to gary’s answer). I’m not saying my proof at the end is better than yours in any way, I’m just showing a slightly alternative way of looking at it.

Define the distance between two non-empty subsets A,BX to be d(A,B)=infaA,bBd(a,b) and write d(x,B) if A={x}.

  1. If BX is arbitrary and non-empty then xd(x,B) is 1-Lipschitz continuous, that is |d(x,B)d(y,B)|d(x,y) for all x,yX.
  2. We have d(x,B)=0 if and only if x¯B.
  3. If d(,A)=d(,B) then ¯A=¯B.
  1. Choose bB such that d(x,b)d(x,B)+ε. Then the triangle inequality yields d(y,B)d(x,B)d(y,b)d(x,b)+εd(y,x)+ε. By symmetry we get |d(x,B)d(y,B)|d(x,y)+ε, and 1. follows because ε was arbitrary. Update: In this closely related answer I show that 1 is in fact the best Lipschitz constant as soon as B isn’t dense. Don’t miss Didier’s answer to the same thread which relies on a useful general fact which is as easy to prove and Zarrax’s answer providing a cleaned-up argument of the one I’m giving here.

  2. Choose bnB with d(x,bn)d(x,B)+1n=1n. Then d(x,bn)0 and hence x¯B. Conversely, if bnx then d(x,bn)0 hence d(x,B)=0.

  3. Immediate from 2.


Let me combine these facts: assume A is compact and B is closed. As d(,B):X[0,) is continuous by 1. above, we conclude from compactness of A that d(,B) assumes its minimum when restricted to A (if you think about how one usually proves this, you’ll find your argument again!). Hence there is aA with the property that d(a,B)d(a,B) for all aA. But if d(a,B)=0 then aB by 2. above, since B=¯B. So either A and B are not disjoint or d(a,B)d(a,B)>0. By choosing δ(0,d(a,B)), we get the claim again.


Finally, if you don’t assume that one among A and B is compact, then the result is false. There was the example A=N and B={n+1n}nN given in the comments, or, a bit more geometrically appealing to me, let A be the x-axis in R2 and B the graph of the function x1x, x0.

Attribution
Source : Link , Question Author : Benji , Answer Author : Troy Woo

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