 Solution

The Tietze extension theorem states: if $X$ is a normal topological space, $A\subseteq X$ is closed, and $f:A\to [-M,M]$ is continuous, then there is a continuous $F:X\to [-M,M]$ with $F{|}_A=f$.

We first prove an approximation lemma. Given continuous $h:A\to [-r,r]$, the closed subsetsare disjoint and closed in $X$. By the Urysohn lemma, there is a continuous $g:X\to [-r/3,r/3]$ equal to $-r/3$ on $A_{-}$ and $r/3$ on $A_{+}$. Then

Starting with $h_0=f$ and $r_0=M$, apply the lemma recursively to obtain continuous $g_j:X\to \mathbb{R}$ such thatwhere $r_j=M(2/3{)}^j$. The seriesconverges uniformly by the Weierstrass test, so its sum is continuous. On $A$, its remainder after $j$ terms is $h_{j+1}$, which tends uniformly to zero; hence $F{|}_A=f$. The bounds also keep $F$ in $[-M,M]$ after the standard endpoint-preserving version of the construction, proving the theorem.