Tietze extension theorem

Statement

Let $X$ be a normal space, let $A \subseteq X$ be closed, and let $f \colon A \to \mathbb{R}$ be continuous for the subspace topology on $A$. Then there is a continuous $F \colon X \to \mathbb{R}$ whose restriction to $A$ is $f$. If also $|f(a)| \le M$ for all $a \in A$, then one can choose $F$ with $|F(x)| \le M$ for all $x \in X$.

Remarks

The proof applies Urysohn's lemma many times and sums a uniformly convergent series of functions. The converse holds: if each continuous function from each closed subset to $\mathbb{R}$ extends, then $X$ is normal. The closedness of $A$ is needed: the function $\sin(1/x)$ on $(0, 1]$ has no continuous extension to $[0, 1]$.

Explanations

L1 human

On a normal space, each continuous real function on a closed set can be extended to the whole space, without losing continuity. You can also keep the same bound on the function.