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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?

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    $\begingroup$ 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. $\endgroup$
    – Thorgott
    Dec 7 '19 at 13:34
  • $\begingroup$ $[0,1]=[0,1/2]\cup(1/2,1]$. One would get a different definition. $\endgroup$ Dec 7 '19 at 13:35
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    $\begingroup$ If we require both sets to be closed, then the definition is equivalent to the original definition. $\endgroup$
    – GEdgar
    Dec 7 '19 at 14:18
  • $\begingroup$ Thank you for these comments! Very helpful. $\endgroup$
    – user489116
    Dec 7 '19 at 20:51
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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.

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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.

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  • $\begingroup$ 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? $\endgroup$
    – user489116
    Dec 7 '19 at 20:58

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