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

1. 1.

   $X$ is normal.

2. 2.

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

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

The present result can be found in [1].