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May 25

sufficient condition for two set to be sepreate-able (topology)

Hi, I look for a sufficient condition that for the next claim to hold;

Let A be a compact subset ${ℝ}^n$
and let B1 and B2 two different connected component. So there are 2 Open subset $O_1,O_2$

1. $O_1\cup O_2\supseteq A$
2. $cl(O_1)\cap cl(O_2)=\mathrm{\varnothing }$
3. $B_1\subseteq O_1$
4. $B_2\subseteq O_2$

Thanks!--Exx8 (talk) 19:01, 25 May 2021 (UTC)

Is there supposed to be some relation between $A$ and the pair $\{B_1,B_2\}$, or are all three just given?  --Lambiam 23:44, 25 May 2021 (UTC)

B1 and B2 are connected component of A.--Exx8 (talk) 05:14, 26 May 2021 (UTC)

No additional conditions are necessary. The statement is true as given.--2406:E003:855:9A01:74B0:C329:6D75:B8CD (talk) 06:49, 26 May 2021 (UTC)

Can you prove it?--Exx8 (talk) 12:52, 26 May 2021 (UTC)

Is it true that the connected components of a compact space are all open? If so, one can take $O_1=B_1,$ $O_2=A\setminus B_1.$  --Lambiam 08:43, 27 May 2021 (UTC)

No. The connected components of the Cantor middle third set are the singletons.2406:E003:855:9A01:6D91:C1FE:E529:AA45 (talk) 00:00, 28 May 2021 (UTC)

@Exx8 Maybe this is equivalent to saying the components are all bounded away from each other- that is, given any two components $B_1,B_2$ there is some $ϵ>0$ such that $d(x,y)>ϵ$ whenever $x\in B_1$ and $y\in B_2$. Staecker (talk) 11:32, 28 May 2021 (UTC)

Not equivalent (consider $\{(x,y)\in {ℝ}^2:|xy|>1\}$) but the implication goes the right direction. --JBL (talk) 13:28, 28 May 2021 (UTC)