# A topological space is disconnected if and only if there exists a non-constant continuous function from the space to the discrete space on two elements

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## Contents

## Statement

Let [ilmath](X,\mathcal{ J })[/ilmath] be a topological space, and let [ilmath](Y,\mathcal{ K })[/ilmath] be a discrete topological space with [ilmath]Y:=\{0,1\}[/ilmath] and [ilmath]\mathcal{K}:=\mathcal{P}(Y)[/ilmath]^{[Note 1]} then^{[1]}:

- [ilmath](X,\mathcal{ J })[/ilmath] is disconnected (ie, not connected)
*if and only if*there exists a non-constant^{[Note 2]},*continuous*function, [ilmath]f:X\rightarrow Y[/ilmath]

## Proof

Grade: C

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The message provided is:

Easy proof to do, exercise on page 87 in Lee's top. manifolds, but it didn't take me very long. Marked as easy

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

- Every continuous map from a non-empty connected space to a discrete space is constant
- A topological space is disconnected if and only if it is homeomorphic to a disjoint union of two or more non-empty topological spaces

## Notes

- ↑ Note: [ilmath]\mathcal{P}(Y)=\mathcal{P}(\{0,1\})=\{\emptyset,\{0\},\{1\},\{0,1\}\}[/ilmath]
- ↑ Recall a map is constant if:
- [ilmath]\forall p,q\in X[f(p)=f(q)][/ilmath]

## References