Tietze extension theorem

Statement
Suppose $$X$$ is a normal space (i.e., a topological space that is T1 and where any two disjoint closed subsets can be separated by disjoint open subsets). Suppose $$A$$ is a closed subset of $$X$$, and $$f:A \to [0,1]$$ is a continuous map. Then, there exists a continuous map $$g:X \to [0,1]$$ such that the restriction of $$g$$ to $$A$$ is $$f$$.

Facts used

 * 1) uses::Urysohn's lemma: This states that for, given two closed subsets $$A,B$$ of a normal space $$X$$, there is a continuous function $$h:X \to [0,1]$$ such that $$h(A) = 0$$ and $$h(B) = 1$$.
 * 2) uses::Closedness is transitive: A closed subset (in the subspace topology) of a closed subset is closed in the whole space.

Proof
Note that since $$[0,1]$$ is homeomorphic to $$[-1,1]$$, it suffices to prove the result replacing $$[0,1]$$ with $$[-1,1]$$. We will also freely use that any closed interval is homeomorphic to $$[0,1]$$, so Urysohn's lemma can be stated replacing $$[0,1]$$ by any closed interval.

Given: A normal space $$X$$. A closed subset $$A$$ of $$X$$. A continuous function $$f:A \to [-1,1]$$.

To prove: There exists a continuous function $$g:X \to [-1,1]$$ such that the restriction of $$g$$ to $$A$$ is $$f$$.

Proof: We write $$f_0 = f$$.

The key part of the proof is the inductive portions (6)-(10). Steps (1)-(5) can be thought of as a special case of these; however, these steps are provided separately so that the full details can be seen in the base case before the general inductive setup is used.