# 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

From Maths

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## Contents

## 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^{[1]}:

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

- [ilmath]\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}][/ilmath] - the "disjoint in [ilmath]A[/ilmath]" condition is perhaps better written as: [ilmath](U\cap A)\cap(V\cap A)=\emptyset[/ilmath]
- In words - "
*there exist two sets open in [ilmath](X,\mathcal{ J })[/ilmath] that are disjoint in [ilmath]A[/ilmath] and non-empty in [ilmath]A[/ilmath] that cover [ilmath]A[/ilmath]*"

- In words - "

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

## Proof

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**This proof has been marked as an page requiring an easy proof**## 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

TODO: Flesh out

## References