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].
References
 1 A. Mukherjea, K. Pothoven, Real and Functional analysis^{}, Plenum press, 1978.
Title  Tietze extension theorem 

Canonical name  TietzeExtensionTheorem 
Date of creation  20130322 13:35:30 
Last modified on  20130322 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 