Difference between revisions of "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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==See also== | ==See also== | ||
* [[Every continuous map from a non-empty connected space to a discrete space is constant]] | * [[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== | ==Notes== | ||
<references group="Note"/> | <references group="Note"/> | ||
Latest revision as of 23:12, 30 September 2016
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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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
