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Let $(X,d)$ be a metric space. Prove that $E$ is disconnected iff there exists two open disjoint sets $A$,$B$ in $X$ such that $E\cap A\neq\emptyset, E\cap B\neq \emptyset$ and $E\subset A\cup B$.

I'm not sure how to begin, so let me just start by pointing some stuff out from the question that I noticed, and my knowledge as of now. (Hopefully it'll be useful.)

  • There is $a\in A, b\in B$ which are both limit points of $E$ (I think)

  • Since $A$ and $B$ are disjoint, $A\cap B=\emptyset$

  • The closure of a set is the "smallest" set which contains its limit points, (so it is a closed set as well).

  • Both $A$ and $B$ are open so they are made up entirely of interior points, so they don't have any points on the boundary.

I know that the definition of separated means that I have $A,B\subset X$, for which $A\cap \bar{B}=\emptyset=\bar{A}\cap B$, where the bar above the set denotes its closure.

And I know that the definition of connected set means for $E\subset X$, $E$ is connected if $E$ is not the union of two non-empty separated sets, so a disconnected set would be for $E\subset X$, $E$ is disconnected if it is the union of two separated sets (which are both non-empty).

Any hints as to how to begin would be appreciated. Thank you.

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  • $\begingroup$ Do you mean the disjoint sets $A$ and $B$ to be both open (or both closed)? $\endgroup$
    – Singhal
    Commented Feb 24, 2014 at 6:47
  • $\begingroup$ My apologies- that's a typo. Let me correct it. Both $A$ and $B$ are open $\endgroup$ Commented Feb 24, 2014 at 6:48
  • $\begingroup$ Do you know Urysohn's lemma? $\endgroup$ Commented Feb 24, 2014 at 6:54
  • $\begingroup$ @EricTowers: I'm afraid I do not. $\endgroup$ Commented Feb 24, 2014 at 6:55
  • $\begingroup$ You say that $E$ is disconnected if it is the union of two disconnected sets. Is $E$ a union of two sets? (You have enough given to answer this.) Are those two sets disjoint? Does disjoint imply separated in a metric space (and/or any other first countable space)? $\endgroup$ Commented Feb 24, 2014 at 6:57

1 Answer 1

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Paraphrased from my answer here: If there are disjoint open sets $A,B$ which intersect $E$ and such that $A\cup B\supseteq E$, then $A\cap E$ and $B\cap E$ form a disconnection of $E$, so $E$ is disconnected. Conversely, suppose that $A'$ and $B'$ are open sets such that $A'\cup B'\supseteq E$ and $A'\cap E\ne\emptyset$, $B'\cap E\ne\emptyset$, $A'\cap B'\cap E=\emptyset$ (by definition of disconnected). Then setting

$$A=\{x:d(x,A'\cap E)<d(x,B'\cap E)\}$$ $$B=\{x:d(x,A'\cap E)>d(x,B'\cap E)\}$$

we have that $A$ and $B$ are open sets (since $d(x,A'\cap E)-d(x,B'\cap E)$ is a continuous function), disjoint, and for any $x\in A'\cap E$, letting $r$ be such that $B(r,x)\subseteq A'$, we have $d(x,A'\cap E)=0$ (of course) and $d(x,B'\cap E)\ge r$ (since $A'\cap B'\cap E=\emptyset$), so $A\ne\emptyset$, and similarly $B\ne\emptyset$.

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