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?
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.
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.
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 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.
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.
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 non-empty.
Many thanks! Indeed my proof made no sense whatsoever.the_sheath wrote:X
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