Why does the definition of (dis-) connectedness use open sets?

My course on metric spaces uses the following definition for connectedness:

$$(X,d)$$ is disconnected if there exist open $$U,V \subset X$$ such that $$X = U \cup V; \ U,V \neq \emptyset$$ and $$U \cap V = \emptyset$$. $$(X,d)$$ is connected if it is not disconnected.

I started playing a bit with this definition and wondered how necessary the requirement of openness on $$U$$ and $$V$$ is. That is would we get an absurd definition if we allowed one set to be closed/clopen?

• If there is an open $U$, such that $\emptyset\neq U\neq X$, then $X=U\cup X\setminus U$. This union is disjoint, $U$ is open, $X\setminus U$ is closed and neither of these two is empty. So if one set were allowed to be closed, almost all spaces would be disconnected. That would not be very interesting. Commented Dec 7, 2019 at 13:34
• $[0,1]=[0,1/2]\cup(1/2,1]$. One would get a different definition. Commented Dec 7, 2019 at 13:35
• If we require both sets to be closed, then the definition is equivalent to the original definition. Commented Dec 7, 2019 at 14:18
– user489116
Commented Dec 7, 2019 at 20:51

The original definition for disconnectedness is that there exists subsets $$A,B$$ of $$X$$ such that $$X=A\cup B$$ and $$\emptyset \neq A \neq X$$, $$\emptyset \neq B \neq X$$ ($$A$$ and $$B$$ are proper, non-empty subsets) and $$A$$ and $$B$$ are separated: $$\overline{A} \cap B = \emptyset = A \cap \overline{B}$$: no point can be in $$A$$ and close to $$B$$ too (or vice versa in $$B$$ and close to $$A$$), a stronger version of being merely disjoint.

The basic idea is that $$X$$ has two pieces that are "apart from each other".

Now, in the above situation, as $$X=A \cup B$$ we in fact have that $$A$$ and $$B$$ are both closed in $$X$$: no point of $$\overline{A}$$ can be in $$B$$, but it has to be in $$A$$ or $$B$$ so must be in $$A$$ and $$A$$ is thus closed, or formally

$$A \subseteq \overline{A} \subseteq X\setminus B \subseteq A \text{ so } \overline{A}=A$$ and likewise for $$B$$.

But as $$A$$ and $$B$$ are each other's complement they are also both open in $$X$$.

So in a disconnecting partition we can ask for either two separated sets, or two disjoint open sets (which are automatically separated) or two disjoint closed sets (also automatically separated). So in a way it doesn't matter, a text has to choose one definition and stick to it.

Open disjoint sets is just one way for a space to be disconnected.
Exercise.
Prove if K,L are not empty, disjoint, closed subsets of S,
and K $$\cup$$ L = S, then S is disconnected.

• Thank you! This should really be a new post, but what's the reason for considering openness in the definition of connectedness in the first place? Is it that that is simply the most elementary notion of topology?
– user489116
Commented Dec 7, 2019 at 20:58