A subset of a topological space is disconnected if and only if it can be covered by two non-empty-in-the-subset and disjoint-in-the-subset sets that are open in the space itself

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Statement

Let [ilmath](X,\mathcal{ J })[/ilmath] be a topological space and let [ilmath]A\in\mathcal{P}(X)[/ilmath] be an arbitrary subset of [ilmath]X[/ilmath], then:

• [ilmath]A[/ilmath] is disconnected subset (ie: not connected) if and only if:
  • [ilmath]\exists U,V\in\mathcal{J}[U\ne\emptyset\wedge V\neq\emptyset\wedge U\cap V=\emptyset\wedge A\subseteq U\cup V][/ilmath] - in words, "there exists a non-empty and disjoint covering of [ilmath]A[/ilmath] by sets open in [ilmath](X,\mathcal{ J })[/ilmath]"

Proof

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Leads to

• A space is a disconnected subset of itself if and only if the space itself is disconnected
  • A space is a connected subset of itself if and only if the space itself is connected - one can be used to prove the other

See also

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TODO: Flesh out

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