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

• The topological space $(A,\mathcal{J}_A)$ (which is $A$ imbued with the subspace topology inherited from $(X,\mathcal{ J })$) is disconnected (the very definition of disconnected subset)

if and only if

• $\exists U,V\in\mathcal{J}[\underbrace{\ U\cap A\ne\emptyset\ }_{U\text{ non-empty in }A}\wedge\underbrace{\ V\cap A\ne\emptyset\ }_{V\text{ non-empty in } A}\wedge\underbrace{\ U\cap V\cap A=\emptyset\ }_{U,\ V\text{ disjoint in }A}\wedge\underbrace{\ A\subseteq U\cup V\ }_{\text{covers }A}]$ - the "disjoint in $A$" condition is perhaps better written as: $(U\cap A)\cap(V\cap A)=\emptyset$
  • In words - "there exist two sets open in $(X,\mathcal{ J })$ that are disjoint in $A$ and non-empty in $A$ that cover $A$"

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TODO: There is a formulation similar this (see p114, Mendelson) that works in terms of closed rather than open sets, link to it!

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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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