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Wikipedia provides several different definitions of connectedness that they say are equivalent. For example, it says that a connected set is not the union of disjoint open sets. While this makes sense, why does it suffice to show that if a set is not the union of open sets, it is connected? (Ex. The union of [1,2] and [3,4] is not connected, but it is not the union of open sets.)

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    $\begingroup$ Those sets are open in the correct subspace topology. $\endgroup$
    – Randall
    Commented May 22, 2020 at 19:17
  • $\begingroup$ If the Wikipedia page didn't say "nonempty open sets" then it needs to be corrected. $\endgroup$
    – bof
    Commented Jul 4, 2020 at 7:04

3 Answers 3

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$[1,2]\cup[3,4]$ is union of two open sets and so it is disconnected. Important is looking carefully the definition.

Definition:

A metric space $X$ is said to disconnected if it is union of two non empty disjoint open sets.
A subset $S\subset X$ is said to be disconnected if it is disconnected as a subspace of $X$.

So $S$ will become disconnected if $S=U\cup V$ for disjoint nonempty open sets $U$ and $V$ in $S$.

Now here if you talk about connectedness of $S=[1,2]\cup[3,4]$ then we have to consider it as a subspace of $\Bbb{R}$. Here $[1,2]=S\cap\left(\displaystyle\frac12,\frac52\right)$ and

$[3,4]= S\cap\left(\displaystyle\frac52,\frac92\right)$ are open sets in $S$. Thus $S$ is union of two non empty disjoint open sets and hence not connected.

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We have to understand that connectedness must be defined in a metric (or topological) space. So, when we say "open" sets, we mean that they are open subject to a defined (metric) topology.

Simply, the connected (space) set $A \subseteq X$ is a set that is contained in a metric (or topological) space and there are not any two disjoint open (which implies the closed too) sets (in $X$) that make a partition for $A$. Namely, for any $U$ and $V$ non-empty open in $X$ and $U \cap V =\phi$, then $U \cup V \neq A$.

In conclusion, a set $A$ is connected if we can not find a partition for A from the open sets (subject to the topology (metric)).

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    $\begingroup$ Somewhere in the sentence beginning "Namely" you need to say that the open sets are nonempty. (You were okay in the preceding sentence because "nonempty" is implicit in "partition".) $\endgroup$
    – bof
    Commented Jul 4, 2020 at 7:03
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What is meant is that the sets should be open in the space itself. In the union of $[1,2]$ and $[3,4]$, each of the intervals is open in the subspace topology, even though they are not open in the ambient topology.

A disconnected space is one that is exactly equal to the disjoint union of two of its nonempty open subsets. For a subspace $A$, it's enough for there to exist two nonempty open sets $U$ and $V$ in the ambient space such that $U\cap A$ and $V\cap A$ are disjoint and $A\subseteq U\cup V$.

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  • $\begingroup$ @bof True, fixed. $\endgroup$ Commented Jul 4, 2020 at 11:45

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