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I am trying to prove that if $X$ is a connected subset of a metric space $M$, then $\bar{X}$ is connected as well, that is, the closure of $X$ is connected.

My proof:

I will do a proof by contradiction. Suppose that $\bar{X}$ is disconnected. Then, there exist open set $U,V \subset \bar{X}$ which are relatively open in $X$, non empty and disjoint. Now, $U' = U \cap X$ and $V' = V \cap X$ are relatively open in $X$. Now, I was told by my book that to obtain the contradiction, we would like to show that $U', V'$ are both non-empty. I do not understand why showing $U', V'$ are both non-empty will lead to a contradiction. Anyone know? Thank you!

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3 Answers 3

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So suppose $X$ is connected and $\overline{X}$ is disconnected, so there are relatively open subsets $U,V$ of $\overline{X}$ that are both non-empty, disjoint and $U \cup V = \overline{X}$. So there are open sets $U',V'$ in $M$ such that $U = U' \cap \overline{X}$ and $V = V' \cap \overline{X}$, by the definition of relative (or subspace) topology.

Claim: $U' \cap X \neq \emptyset$. This follows as otherwise every point of $U'$ is not in $\overline{X}$, but we know that all points of $\emptyset \neq U \subset U'$ are in $\overline{X}$... Symmetrically, $V' \cap X \neq \emptyset$ as well.

But then $U' \cap X$ and $V' \cap X$ are disjoint, non-empty (see above) and cover $X$ (as $X = U \cup V \subset U' \cup V'$). This contradicts that $X$ is connected.

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  • $\begingroup$ When you write that otherwise every point of $U'$ is not in $\overline{X}$, does this follow because they also cant be elements of the boundary of $X$? This would be because they are points of an open set, so we should be able to fit a neighbourhood completely inside the boundary, but evidently this isn't possible? Am I right in thinking this breaks down if we think about the closure of a subspace of a general topological space? $\endgroup$ Mar 6, 2022 at 1:24
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    $\begingroup$ @AndreyYanyuk no, that’s not the reason. In any space, point is in the closure of a set iff every neighbourhood of that point intersects the set. That’s what I use. $\endgroup$ Mar 6, 2022 at 7:48
  • $\begingroup$ Ah perfect thanks very much! $\endgroup$ Mar 6, 2022 at 14:47
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If $U',V'$ are both non-empty, then this would imply that $X$ is not connected, which contradicts the hypotheses.

The relevant result is that if an open set intersects the closure of a set, then it intersects the set itself. This is straightforward to establish.

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The first line of the proof by contradiction should be "suppose $\bar{X}$ is disconnected and $X$ is connected. Then when you show $U^\prime$ and $V^\prime$ are non-empty, you have a connected set containing two disjoint non-empty open sets, which is a contradiction (per the definition of connected).

The harder part, or course, is showing that $U^\prime$ and $V^\prime$ are indeed non-empty.

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  • $\begingroup$ I see, so basically the contradiction comes from $U',V'$ being disjoint, open, nonempty subset of $X$? So is it fair to say that we first "get" $U',V'$ from $\bar{X}$, then show that $U',V'$ actually are subsets of $X$ that imply $X$ is disconnected? Thanks! $\endgroup$
    – user123276
    Jun 10, 2014 at 5:33

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