Simple (?) topology problem
Simple (?) topology problem
I am trying to solve this topology problem that is intuitively completely obvious and yet I have no idea how to proceed in order to formally prove it. I have been staring at it for hours now. Perhaps someone could give me a hint on how to get started, since I am really stuck here.
Let X be a metric space. Let A be a compact subset of X. Let B be a closed subset of X. Assume that for each epsilon > 0 there exist x in A and y in B such that d(x,y) < epsilon. Prove that A intersect B is nonempty.
Let X be a metric space. Let A be a compact subset of X. Let B be a closed subset of X. Assume that for each epsilon > 0 there exist x in A and y in B such that d(x,y) < epsilon. Prove that A intersect B is nonempty.
Re: Simple (?) topology problem
If you construct a sequence of epsilons decreasing to zero, you can also find a sequence in B that converges to a point that is in B because B is closed. Why is that point also in A?
Re: Simple (?) topology problem
You want to find the point in A first, by using compactness.poorasian wrote:If you construct a sequence of epsilons decreasing to zero, you can also find a sequence in B that converges to a point that is in B because B is closed. Why is that point also in A?
Re: Simple (?) topology problem
Assume the intersection is empty. Use the fact that B is closed and disjoint from A and A compact to find a certain finite covering of A. Derive a contradiction.
Re: Simple (?) topology problem
Hi guys,
Thanks for your replies. I attempted to prove it by using some of your suggestions, but I am far from sure it is correct now. Suggestions/corrections are more than welcome! (Please forgive me for my awkward paraphrasing of my LaTeX code.)
Let (epsilon_n) be a sequence in R that converges to 0. Now construct a sequence (x_n) such that x_n is in the closure of B for all n in N and dist(x_n, A) = epsilon_n. Because (epsilon_n) converges to 0, (x_n) converges to 0 and there exists a limit x_n in the closure of B such that dist(x_n, A) = 0. Now it holds that x_n is in the closure of A and x_n is in B because B is closed. A is compact, hence sequentially compact. This means that every sequence in X has a convergent subsequence in A and therefore x_n in A. Conclusion: x_n is in the intersection of A and B which is nonempty.
Thanks for your replies. I attempted to prove it by using some of your suggestions, but I am far from sure it is correct now. Suggestions/corrections are more than welcome! (Please forgive me for my awkward paraphrasing of my LaTeX code.)
Let (epsilon_n) be a sequence in R that converges to 0. Now construct a sequence (x_n) such that x_n is in the closure of B for all n in N and dist(x_n, A) = epsilon_n. Because (epsilon_n) converges to 0, (x_n) converges to 0 and there exists a limit x_n in the closure of B such that dist(x_n, A) = 0. Now it holds that x_n is in the closure of A and x_n is in B because B is closed. A is compact, hence sequentially compact. This means that every sequence in X has a convergent subsequence in A and therefore x_n in A. Conclusion: x_n is in the intersection of A and B which is nonempty.

 Posts: 3
 Joined: Thu Mar 13, 2014 12:38 pm
Re: Simple (?) topology problem
Hi Kleene,
I'm not sure I understand your previous post. Why does (epsilon_n) converges to 0 imply (x_n) converges to 0? Here's another possible way: We know that d(.,B) is a continuous function on a metric space X for any subset B of X. Since A is compact, this function should attain a minimum, and this minimum should of course be 0. Now, if x is a point of A at which d(x,B)=0, then x should also be in B since B is closed.
I'm not sure I understand your previous post. Why does (epsilon_n) converges to 0 imply (x_n) converges to 0? Here's another possible way: We know that d(.,B) is a continuous function on a metric space X for any subset B of X. Since A is compact, this function should attain a minimum, and this minimum should of course be 0. Now, if x is a point of A at which d(x,B)=0, then x should also be in B since B is closed.
Re: Simple (?) topology problem
Why are you only working in R? The statement of your problem regards an arbitrary metric space.
Re: Simple (?) topology problem
He's not working in R, but the codomain of any metric is R, so that's why he's choosing a sequence of epsilons in R. He's still using an arbitrary metric d elsewhere.dasgut wrote:Why are you only working in R? The statement of your problem regards an arbitrary metric space.
Re: Simple (?) topology problem
You are right. That's my bad. I meant to say that (epsilon_n) converges to 0 implies that dist(x_n, B) converges to 0 and hence (x_n) converges (not necessarily to 0). Does that make more sense to you.akbar lipstick wrote:Hi Kleene,
I'm not sure I understand your previous post. Why does (epsilon_n) converges to 0 imply (x_n) converges to 0? Here's another possible way: We know that d(.,B) is a continuous function on a metric space X for any subset B of X. Since A is compact, this function should attain a minimum, and this minimum should of course be 0. Now, if x is a point of A at which d(x,B)=0, then x should also be in B since B is closed.
Your proof seems consistent. I am not sure, however, why the attained minimum would have to be 0. Would you mind explaining that?
Thanks!
Re: Simple (?) topology problem
Kleene wrote:You are right. That's my bad. I meant to say that (epsilon_n) converges to 0 implies that dist(x_n, B) converges to 0 and hence (x_n) converges (not necessarily to 0). Does that make more sense to you.akbar lipstick wrote:Hi Kleene,
I'm not sure I understand your previous post. Why does (epsilon_n) converges to 0 imply (x_n) converges to 0? Here's another possible way: We know that d(.,B) is a continuous function on a metric space X for any subset B of X. Since A is compact, this function should attain a minimum, and this minimum should of course be 0. Now, if x is a point of A at which d(x,B)=0, then x should also be in B since B is closed.
Your proof seems consistent. I am not sure, however, why the attained minimum would have to be 0. Would you mind explaining that?
Thanks!
Kleene wrote:Assume that for each epsilon > 0 there exist x in A and y in B such that d(x,y) < epsilon.
Re: Simple (?) topology problem
I was just working this out and this is what I figured. Thanks.Ryker wrote:Kleene wrote:You are right. That's my bad. I meant to say that (epsilon_n) converges to 0 implies that dist(x_n, B) converges to 0 and hence (x_n) converges (not necessarily to 0). Does that make more sense to you.akbar lipstick wrote:Hi Kleene,
I'm not sure I understand your previous post. Why does (epsilon_n) converges to 0 imply (x_n) converges to 0? Here's another possible way: We know that d(.,B) is a continuous function on a metric space X for any subset B of X. Since A is compact, this function should attain a minimum, and this minimum should of course be 0. Now, if x is a point of A at which d(x,B)=0, then x should also be in B since B is closed.
Your proof seems consistent. I am not sure, however, why the attained minimum would have to be 0. Would you mind explaining that?
Thanks!Kleene wrote:Assume that for each epsilon > 0 there exist x in A and y in B such that d(x,y) < epsilon.
I would still like to know what flaws are in my other proof.
Re: Simple (?) topology problem
Everyone is confused by you writing (x_n) converges to 0. Which it doesn't, so perhaps you meant something else. Perhaps you meant the sequence d(x_n, A) converges to 0? In any case, why would that necessarily imply x_n converges? Because I can think of counterexamples.Kleene wrote:I was just working this out and this is what I figured. Thanks.
I would still like to know what flaws are in my other proof.

 Posts: 10
 Joined: Sun Feb 23, 2014 12:25 pm
Re: Simple (?) topology problem
 B is closed. All points in the closure of B are already contained in B. To choose points in the closure of B is redundant.Kleene wrote:Hi guys,
Thanks for your replies. I attempted to prove it by using some of your suggestions, but I am far from sure it is correct now. Suggestions/corrections are more than welcome! (Please forgive me for my awkward paraphrasing of my LaTeX code.)
Let (epsilon_n) be a sequence in R that converges to 0. Now construct a sequence (x_n) such that x_n is in the closure of B for all n in N and dist(x_n, A) = epsilon_n. Because (epsilon_n) converges to 0, (x_n) converges to 0 and there exists a limit x_n in the closure of B such that dist(x_n, A) = 0. Now it holds that x_n is in the closure of A and x_n is in B because B is closed. A is compact, hence sequentially compact. This means that every sequence in X has a convergent subsequence in A and therefore x_n in A. Conclusion: x_n is in the intersection of A and B which is nonempty.
 There may not exist an x_n such that dist(x_n,A) = epsilon_n. Consider the possibility of B being a single point. You want to use "<" or "<=".
 Why does the sequence x_n converge to anything? All we know at this point is that it gets closer and closer to the set A. For all you know, each of the x_n is within epsilon_n of a different point of A, and may not converge to any point of A or any point at all. I'm confused at the point where you say the sequence x_n converges, because at that point, you don't use that A is compact at all. Let A be the integers (I know that isn't a compact set) and let be be the set {n+1/n  n \in Z } and let {x_k} = {k+1/k} where k is {1,2,3}... Clearly {x_k} is in B, but does not converge, nor does any subsequence converge. But it fulfills all of your criteria up to that point. Why can't I make a similar argument if A is compact?
 If x_n converged, why is it contained in the closure of A? I mean, if A was a single point, it would be compact, and you can easily construct a sequence x_n in B to converge to that point in A by the premises given, but x_n would not necessarily be in the closure of A, since A would be closed. Why does it matter that x_n is in the closure of A?
 Not every sequence in X has a convergent subsequence in A. Let X be R, let A be [0,1]. The sequence {42,42,42,42,...} is not contained in A, nor is any of its subsequences. That A is sequentially compact means that every sequence in A contains a convergent subsequence. But you haven't shown that x_n is in A, so that step makes no sense. And even if you have, that step is completely unnecessary, since it would be obvious from the fact that the sequences are contained in both sets that their intersection is nonempty.
Re: Simple (?) topology problem
Many thanks! Indeed my proof made no sense whatsoever.the_sheath wrote:X