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I have this problem:

If $X$ is a metric space, and if $A,B\subseteq X$ then the distance between $A$ to $B$ is $\mathtt d(A,B):=\inf \{d_X(x,y)| x\in A,y\in B\}$. Show that $\mathtt d(A,B)>0$ if $A\cap B=\varnothing$, $A$ is compact and $B$ is closed. What happen or if $B$ is open instead of closed? ans if $A$ is only closed instead of compact?

For the first claim, suppose that $\mathtt d(A,B)\not\gt 0\rightarrow\mathtt d(A,B)=0$, so $\inf \{d_X(x,y)| x\in A,y\in B\}=0$, then there's at least one $x\in A$ such that this happens, now we take the $\mathtt d(x,B):=\inf_{z\in B}d_X(x,z)=0$, I already proved that if that last distance is $0$ and the set is closed then $x\in B$, hence $x\in A\lor x\in B$ wich is a contradiction.

For the second claim, I'm not so sure, I think that in this case the $\inf$ doesn't exists, if we suppose that $\mathtt d(A,B)=\beta>0$ then doing something similar to the above, there's a $y_1\in B$ such that $\mathtt d(y_1,A)=\beta$, on the other hand since $B$ is open then $\exists\; \varepsilon_1>0$ such that $B_X(y_1,\varepsilon_1)\subseteq B$, so there's a $y_2\in B_X(y_1,\varepsilon_1)$ such that for some $0<\varepsilon_2<\varepsilon_1$ then $\mathtt d(y_2,A)+\varepsilon_2=\beta$, then $\mathtt d(y_2,A)<\mathtt d(y_1,A)$, so there is a smaller number than $\beta$. I think I'm taking to much liberty with the $y_2$.

For the third I belive that you just have to give a counterexample, however I'm having trouble finding it. I was thinking maybe something like a line in $\Bbb R^2$ that gets really close to another line, like asymptotically, but I don't know if it'll really work with any metric space.

Ana Galois
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  • For the "what if" parts, all you need is to give counterexamples. Under the conventional definition of line, your two lines idea will not work. But a small modification will, such as one set the $x$-axis, the other the curve $y=1/x$. – André Nicolas Sep 12 '13 at 01:40
  • @AndréNicolas Do I have to prove that $x$-axis and $1/x$ hyperbola are closed? When a problem says "Prove that in general blah doesn't happen if bleh happens" you just have to give a counterexample? – Ana Galois Sep 12 '13 at 03:02
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    Well, the fact that they are closed is pretty obvious, but you should at least say that they are closed. Unfortunately, I cannot give you a general answer. It depends on the level of detail demanded by the instructor. – André Nicolas Sep 12 '13 at 03:12
  • Yeah, I understand. But now, with the counterexample for the 2nd claim, I thought maybe taking $X=\Bbb R$, $A=[0,1]$, $B=(2,3)$, then the infimum distance is 1, but there isn't any elements in A and B that satisfaces that, is it ok? – Ana Galois Sep 12 '13 at 03:25
  • I think you want say $A=[0,1]$ and $B=(1,2)$. – André Nicolas Sep 12 '13 at 03:27
  • Unfortunately, your argument for the first part is at best very incomplete. You have not even used compactness. I am reluctant to write an answer, since this question has been asked and answered several times on MSE. Here is one place. There are others even closer. You are probably much better than I am at searching. – André Nicolas Sep 12 '13 at 03:36
  • that is better, thanks. I know! That's why I don't feel well about what I wrote, obviously somewhere we have to use compactness, and I found this http://math.stackexchange.com/questions/109167/if-a-is-compact-and-b-is-closed-show-da-b-is-achieved , however they end up proving this when both sets are compact, I don't understan why they do that – Ana Galois Sep 12 '13 at 03:45
  • Typing mathematics of any length in comments is painful, so I have given a fairly detailed sketch of an answer for the first part. – André Nicolas Sep 12 '13 at 04:00
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    See also: A and B disjoint, A compact, and B closed implies there is positive distance between both sets. – Martin Sleziak Oct 23 '17 at 13:52

1 Answers1

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We have discussed the last two parts in comments. For the first part, suppose $d(A,B)=0$. Then there is a sequence $(a_n,b_n)$ of ordered pairs, where $a_n\in A$ and $b_n\in B$, such that $\lim_{n\to\infty}d(a_n,b_n)=0$. Because $A$ is compact, the sequence $(a_n)$ has a convergent subsequence $(a_{n_i})$ with limit $a\in A$. Now consider the sequence $(b_{n_i})$. By using the Triangle Inequality, we can show that it has limit $a$. Since $B$ is closed, we have $a\in B$, contradicting the fact that $A\cap B=\emptyset$.

André Nicolas
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