Stone–Weierstrass theorem

Remarks

The idea of the proof: the closure $\bar{A}$ is again an algebra. By the Weierstrass approximation theorem, $\abs{t}$ is a uniform limit of polynomials on each interval, so $f \in \bar{A}$ implies $\abs{f} \in \bar{A}$. Hence $\bar{A}$ is closed under $\max$ and $\min$. A compactness argument then approximates any continuous function by finite maxima and minima of functions in $\bar{A}$.

Examples: polynomials in $n$ variables are dense in $C(K, \R)$ for each compact $K \subset \R^n$. Trigonometric polynomials are dense in the continuous $2\pi$-periodic functions.

Both hypotheses are needed. On $[-1, 1]$, the even polynomials form an algebra with constants, but it does not separate $1$ from $-1$, and $x$ is not a uniform limit of even functions. The functions $f$ with $f(0) = 0$ separate points but miss the constants.

For complex functions, $A$ must also be closed under complex conjugation. On the unit circle, the polynomials in $z$ separate points but are not dense: $\bar{z}$ is not a uniform limit of them.

Going deeper

For a student

The theorem is the standard way to prove that a class of functions is dense in $C(X)$. It gives density of polynomials, of trigonometric polynomials, and of sums of products $f(x) g(y)$ in $C(X \times Y)$. In the theory of C*-algebras it is the main tool behind the commutative Gelfand–Naimark theorem.