# Difference between revisions of "Tietze Extension Theorem"

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+ | ==Theorem== | ||

+ | Let $(X,\tau)$ be a [[normal]] [[topological space]], let $A \subset X$ be [[closed]], and let $f \colon A \rightarrow \mathbb{R}$ be a [[continuous]] [[function]] (where $\mathbb{R}$ is equipped with the [[Euclidean topology]]). Then there exists a continuous function $F \colon X \rightarrow \mathbb{R}$ such that for all $a \in A$, $F(a)=f(a)$. | ||

− | == | + | ==Proof== |

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− | == | + | == See Also == |

* [[Urysohn Lemma]] | * [[Urysohn Lemma]] | ||

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* [[Urysohn Metrization Theorem]] | * [[Urysohn Metrization Theorem]] | ||

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[[Category:General Topology]] | [[Category:General Topology]] | ||

+ | [[Category:Theorem]] | ||

+ | [[Category:Unproven]] |

## Latest revision as of 02:03, 24 December 2018

## Theorem

Let $(X,\tau)$ be a normal topological space, let $A \subset X$ be closed, and let $f \colon A \rightarrow \mathbb{R}$ be a continuous function (where $\mathbb{R}$ is equipped with the Euclidean topology). Then there exists a continuous function $F \colon X \rightarrow \mathbb{R}$ such that for all $a \in A$, $F(a)=f(a)$.