# 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. Dec 7 '19 at 13:34
• $[0,1]=[0,1/2]\cup(1/2,1]$. One would get a different definition. Dec 7 '19 at 13:35
• If we require both sets to be closed, then the definition is equivalent to the original definition. Dec 7 '19 at 14:18
– user489116
Dec 7 '19 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
Dec 7 '19 at 20:58