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The real Stone-Weierstrass theorem states that if $X$ is a compact Hausdorff space and $A\subseteq C(X,\mathbb{R})$ is a subalgebra containing the constants and separating points of $X$, then $A$ is dense in $C(X,\mathbb{R})$ in the uniform norm.

The Tietze extension theorem states that if $X$ is a normal topological space, $F\subseteq X$ is closed, and $f:F\to \mathbb{R}$ is continuous, then $f$ extends to a continuous function on $X$. In its bounded form, an $f$ taking values in $[-M,M]$ has an extension with the same range bound. The underlying separation step is the Urysohn lemma.

The Urysohn lemma states that if $A$ and $B$ are disjoint closed subsets of a normal space $X$, there is a continuous $u:X\to [0,1]$ with $u=0$ on $A$ and $u=1$ on $B$.

The Tietze extension theorem states that if $A$ is a closed subset of a normal space $X$, then every continuous $f:A\to \mathbb{R}$ has a continuous extension $F:X\to \mathbb{R}$. In the bounded form, if $f(A)\subseteq [a,b]$, the extension can be chosen with $F(X)\subseteq [a,b]$.

If $X$ is compact and $f:X\to \mathbb{R}$ is continuous, then $f(X)$ is compact by the continuous image of a compact space theorem. Every compact subset of $\mathbb{R}$ is bounded, so $f$ is bounded.

Conversely, suppose the metric space $X$ is not compact. For metric spaces, compactness is equivalent to sequential compactness, so there is a sequence of distinct points $(x_n)$ having no convergent subsequence. The setis closed: an accumulation point would supply a convergent subsequence. It is also discrete for the same reason. Consequently the functionis continuous. By part (a), $X$ is normal, and the Tietze extension theorem extends $g$ to a continuous $G:X\to \mathbb{R}$. Since $G(x_n)=n$, this extension is unbounded. Therefore, if every continuous real-valued function on $X$ is bounded, $X$ must be compact. This is the bounded-continuous-function characterization of compact metric spaces.

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.