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A compact Hausdorff space is both compact and Hausdorff. Such a space is normal, and every real-valued continuous function on it is bounded.

Let $F,G$ be disjoint closed subsets of $K$. They are compact. For each $x\in F,y\in G$, the Hausdorff space condition gives disjoint open neighbourhoods $U_{xy},V_{xy}$. For fixed $x$, finitely many $V_{xy}$ cover $G$; intersect the corresponding $U_{xy}$ and take the union of those $V_{xy}$ to obtain disjoint open sets $U_x\ni x$ and $V_x\supset G$. Finitely many $U_x$ cover $F$. Their union and the intersection of the corresponding $V_x$ are disjoint open neighbourhoods of $F,G$. The case of an empty set is immediate. Hence $K$ is a normal topological space.

Urysohn lemma states that in a normal topological space satisfying $T_1$, disjoint closed sets $F,G$ admit a continuous function $h:K\to [0,1]$ with $h=0$ on $F$ and $h=1$ on $G$.

Let $A,B$ be disjoint closed subsets of the metric space $(X,d)$; the cases where one is empty are immediate. The distance functionsare continuous, because each is 1-Lipschitz. Their sum is strictly positive: if both distances vanished, closedness would put $x$ in $A\cap B$. Thereforeis continuous, equals zero on $A$, and equals one on $B$. The setsare disjoint open neighbourhoods of $A$ and $B$. Thus every metric space is a normal topological space; this is the normality of every metric space.

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.