Tietze extension theorem

Let $X$ be a topological space. Then the following are equivalent:

1. $X$ is normal.

2. If $A$ is a closed subset in $X$, and $f:A\to [-1,1]$ is a continuous function, then $f$ has a continuous to all of $X$. (In other words, there is a continuous function $f^{\ast }:X\to [-1,1]$ such that $f$ and $f^{\ast }$ coincide on $A$.)

Remark: If $X$ and $A$ are as above, and $f:A\to (-1,1)$ is a continuous function, then $f$ has a continuous to all of $X$.

The present result can be found in [1].

Title	Tietze extension theorem
Canonical name	TietzeExtensionTheorem
Date of creation	2013-03-22 13:35:30
Last modified on	2013-03-22 13:35:30
Owner	matte (1858)
Last modified by	matte (1858)
Numerical id	5
Author	matte (1858)
Entry type	Theorem
Classification	msc 54D15
Related topic	ApplicationsOfUrysohnsLemmaToLocallyCompactHausdorffSpaces