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 acontinuous 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].