# Difficulty with Definition of Disconnectedness w.r.t. closed sets

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 $$E$$.

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.

• 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? Mar 4 '21 at 14:54
• $A \cap E$ 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 Mar 4 '21 at 14:57
• 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? Mar 4 '21 at 16:37
• 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? Mar 4 '21 at 16:40
• @strawberry-sunshine yes, closed in $E$ not in $M$, mind you. Mar 4 '21 at 16:41

$$E\subset M$$ 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.

$$E \backslash A$$ 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.