Tietze's extension theorem says that normal spaces have enough room to extend bounded continuous real-valued functions from closed subspaces.
Extension from closed sets
The theorem strengthens Urysohn's lemma from separating two closed sets to extending an entire function.
Theorem. Let \(X\) be normal, let \(A\subseteq X\) be closed, and let \(f:A\to[-1,1]\) be continuous. Then there exists a continuous \(F:X\to[-1,1]\) such that \(F|_A=f\).
Proof. The proof is the approximation construction below. Urysohn's lemma first extends a coarse three-level approximation to \(f\), reducing the remaining error by a factor \(2/3\). Iterating produces continuous functions \(g_n\) with uniformly summable norms, and the uniform limit \(F=\sum_n g_n\) is continuous and agrees with \(f\) on \(A\).
Remark. The closedness of \(A\) is essential: continuous functions on dense nonclosed subsets may force incompatible limiting values.
Approximation step
The proof repeatedly extends coarse sign information and reduces the remaining error.
Lemma. If \(f:A\to[-r,r]\) is continuous on a closed subset \(A\) of a normal space \(X\), then there is a continuous \(g:X\to[-r/3,r/3]\) such that \(|f(a)-g(a)|\le 2r/3\) for all \(a\in A\).
Proof. Let \(A_-=\{a:f(a)\le -r/3\}\) and \(A_+=\{a:f(a)\ge r/3\}\). These are disjoint closed subsets of \(A\), hence closed in \(X\). By Urysohn's lemma, choose \(h:X\to[0,1]\) with \(h(A_-)=0\) and \(h(A_+)=1\). Set \(g=(2r/3)h-r/3\). Then \(g=-r/3\) on \(A_-\), \(g=r/3\) on \(A_+\), and everywhere \(g\in[-r/3,r/3]\), giving the stated error bound.
Uniform convergence construction
The error reduction gives a uniformly convergent series of continuous functions.
Proof. Start with \(f_0=f\) and \(r_0=1\). Apply the lemma to obtain \(g_0:X\to[-1/3,1/3]\) with residual \(f_1=f_0-g_0|_A\) bounded by \(2/3\). Repeat with \(r_n=(2/3)^n\), obtaining \(g_n\) bounded by \(r_n/3\) and residual bounded by \(r_{n+1}\). The series \(F=\sum_{n=0}^\infty g_n\) converges uniformly by the Weierstrass test, hence is continuous. On \(A\), the partial sums differ from \(f\) by a residual tending uniformly to \(0\), so \(F|_A=f\). The range can be clipped to \([-1,1]\) by composing with the continuous retraction \(t\mapsto\max(-1,\min(t,1))\).
Corollary. The theorem holds for any bounded interval \([a,b]\) by an affine change of scale.
Proof. Map \([a,b]\) linearly onto \([-1,1]\), extend there, and map back.
Relation with normality
Tietze extension is not merely a consequence of normality; it characterizes it among \(T_1\) spaces.
Proposition. For a \(T_1\) space \(X\), Tietze extension for all continuous maps from closed subsets into \([-1,1]\) implies normality.
Proof. Let \(A,B\subseteq X\) be disjoint closed sets. Define \(f:A\cup B\to[-1,1]\) by \(f(A)=-1\) and \(f(B)=1\). This is continuous because \(A\) and \(B\) are closed in \(A\cup B\). Extend \(f\) to \(F:X\to[-1,1]\). Then \(F^{-1}((-2,-1/2))\) and \(F^{-1}((1/2,2))\) are disjoint open neighborhoods of \(A\) and \(B\).