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Here's what I know:
When is a (metric) space disconnected?

Consider a metric space $(M,d)$. $M$ is disconnected if there exist non-empty disjoint open subsets $A,B \subset M$ such that $A\cup B = M$. $\{A,B\}$ is called a disconnection of $M$.

When is a subset of $M$ disconnected?$\color{red}{^1}$

Consider $E\subset M$. $E$ is disconnected if there exists non-empty disjoint open subsets $A,B \subset M$ such that $E\subset A\cup B$, $A\cap E\neq \varnothing$ and $B\cap E\neq \varnothing$.

I have seen that using the first definition, people often say that $M$ is disconnected if there exist non-empty disjoint closed sets such that $A\cup B = M$. This makes sense here, since $A^c = B$ and $B^c = A$, so if $A,B$ are open, they are also closed.

What about the second definition though, i.e. when is $E$ disconnected? Can we come up with something in terms of closed sets here?

My problem is that in the second definition, $A^c$ is not necessarily $B$ - so nothing useful comes out of that apparently. Could someone help me out here, and better clarify notions of connectedness/disconnectedness for me? Thank you very much.


Footnote:
$\color{red}{1.}$ In Carothers' Real Analysis, this characterization of disconnectedness of subsets of $M$ has been motivated by the relative metric, and notions of open sets in $E$.

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A subset $E \subseteq M$ is disconnected exactly when $(E,d)$ in the relative metric (or as a topologist would say, the subspace topology) is disconnected.

We can also use the same reformulation as you have given but with closed sets instead. So

$E$ is disconnected if there exists non-empty disjoint closed subsets $A,B \subset M$ such that $E\subset A\cup B$, $A\cap E\neq \varnothing$ and $B\cap E\neq \varnothing$.

The reason that this works is that $A \cap E$ and $B \cap E$ are then open and closed in $(E,d)$ just as in the open $A$,$B$ case.

Sidenote: for general (non-metric) spaces we don't demand that $A \cap B = \emptyset$ but $A \cap B \cap E = \emptyset$, to avoid weird examples). So the sets need only be disjoint on $E$, not in the whole space.

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  • $\begingroup$ How can I prove that the reformulation with closed sets is actually true? In the case of the first definition, the proof is easy. In case of this (second), is there a nice way to reformulate it using the open sets version? $\endgroup$ Mar 4 '21 at 14:54
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    $\begingroup$ $A \cap E$ and $B \cap E$ are each other's complement in $E$. So in the open case, they are both open and closed in $(E,d)$ and in the closed case the same holds, entirely symmetrically... @strawberry-sunshine $\endgroup$ Mar 4 '21 at 14:57
  • $\begingroup$ Well just to confirm: Suppose $E\subset M$, and $A$ is open in $E$. That already means that $A\subset E$, right? Also, the complement of $A$ in $E$ should be $E\setminus A$ - that would be closed right? $\endgroup$ Mar 4 '21 at 16:37
  • $\begingroup$ Also, I'm not too sure about the side-note. Carothers' Real Analysis, Pg. $80$ says otherwise. It seems to demand that $A,B$ are disjoint in $M$, i.e. $A\cap B = \varnothing$. Are you sure? $\endgroup$ Mar 4 '21 at 16:40
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    $\begingroup$ @strawberry-sunshine yes, closed in $E$ not in $M$, mind you. $\endgroup$ Mar 4 '21 at 16:41
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$E\subset M$ is disconnected if there exist two non empty sets $A,B$ such that $A\cap \bar B$ and $\bar A\cap B$ are empty such that $E=A\cup B$. Here, $\bar X =X\cup X'$, where $X'$ is the set of all limit points of $X$.

If $A$ and $B$ are closed, then $A'\subseteq A, B'\subseteq B $, so your definition follows from here.

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$E \backslash A$ is closed IN $E$ under the subspace topology since $E \backslash A = A^c \cap E$. Similarly with $B$. In particular, the same idea works for $E \subseteq M$, you just have to be careful with the subspace topology.

Another very useful definition for disconnectedness goes as follows: $X$ is disconnected if there exists a nonempty, proper subset of $X$ that is both open and closed.

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