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Prove that for a compact metric space $X,$ [$X$ cannot be written as $X=A\cup B$ for nonempty subsets $A$ and $B$ st $d(A,B)>0$] iff $X$ is connected

The question states that one direction doesn't need compactness. I'm fairly confident it's the ==> direction but I'm not sure how to proceed with the proof:

Attempt: ==> For sake of contradiction, assume $X$ is not connected,then $X=A\cup B$ where $A$, $B$ are non-empty, separated sets. Since $d(A,B)\ge0$ as the distance function is always non-negative, then $ \inf{d(A,B)}=0$. However, no $a\in A$ is limit point of $B$ and no $b\in B$ is limit point of $A$

And I'm stuck.

<== For sake of contradiction, assume $X=A\cup B$ such that $d(A,B)>0$. Then since $X$ is connected and $A$ and $B$ are not separated sets, wlog, there must exist $\alpha \in A$ that is limit point of B. However, $\inf{d(\alpha,B)}=0,$ giving us our contradiction.

Edit: Fixing formatting, hold on.

Edit 2: hopefully fixed

Anne Bauval
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Gnolius
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  • Thanks for the help, hopefully everything looks okay now. For "separated", I mean $A^- \cap B = A \cap B^- = \varnothing$. Also, not too sure what you mean by closed or open. Still thinking about the $d(A,B)>0$ but I think I see what you mean, might need some time – Gnolius Jan 09 '23 at 15:00
  • Your definition of "separated" is unusual to me, as well as your definition of "not connected" which uses it. The common definition of not connected is: union of two non-empty complementary subsets which are open or (equivalently, since they are complement of eachother) closed. – Anne Bauval Jan 09 '23 at 15:08
  • In ==>, your arguments do not prove your claim $ d(A,B)=0$ (and there should be no "inf"). And the "However ..." which follows does not conclude to a contradiction. I agree with your <==. – Anne Bauval Jan 09 '23 at 15:16
  • Was trying to say that $d(A,B)>=0$ because of it being a distance function and $d(A,B)<=0$ because we the question tells us that $d(A,B)$ cannot be positive and thus $d(A,B)=0$ Everything from the however onward was just me guessing what would be the next steps, but it probably leads to nowhere after looking at your solution – Gnolius Jan 09 '23 at 15:28
  • Oh now I see! You assume (fsoc ;-)) both $X$ not connected and "$X$ cannot be written $A\cup B$ with $A,B$ nonempty and $d(A,B)>0$"! OK – Anne Bauval Jan 09 '23 at 15:34
  • Your "However..." was then not so far, there was just a compacity argument missing. Note that it is ==> which needs compacity. In our <== we did not need it. – Anne Bauval Jan 09 '23 at 15:58

1 Answers1

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  • $\Rightarrow.$ Assume by contradiction that $X$ is not connected, i.e. $X=A\cup B$ where $A,B$ are non-empty disjoint closed subsets. Then $A$ is compact (since closed in a compact Hausdorff space) and disjoint from the closed subset $B,$ hence $d(A,B)>0.$
  • $\Leftarrow.$ Assume that $X=A\cup B$ with $A,B$ non-empty and $d(A,B)>0.$ Then $X=\bar A\cup\bar B$ and $d(\bar A,\bar B)=d(A,B)>0$ hence $\bar A\cap\bar B=\varnothing,$ which proves that $X$ is not connected.
Anne Bauval
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  • I'm having trouble understanding how $A$ being compact and disjoint from closed subset $B$ would give us d(A,B)>0. I do see some other proofs online using sequences but the book I'm following has not used any sequences and limits yet. Is there another way? – Gnolius Jan 09 '23 at 15:31
  • I added a link, but a simpler way than sequences is to say: by continuity, $a\mapsto d(a,B)$ attains its inf on the compact $A$, for some $a_0\in A,$ hence this inf, $d(A,B)=d(a_0,B),$ cannot be $0$ because $B$ is closed and $a_0\notin B.$ – Anne Bauval Jan 09 '23 at 15:54
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    Thanks for your time and patience, pretty sure I got it now – Gnolius Jan 09 '23 at 15:59