Trigonometric polynomials are uniformly dense on the unit circle

Given: The unit circle $\mathbb{T}$ with the subspace topology from the usual complex metric, and the algebra $T$ of complex trigonometric polynomials on it.

Every unital point-separating self-adjoint complex function algebra on a compact Hausdorff space is uniformly dense in the full complex continuous-function space (Complex Stone–Weierstrass dichotomy for separating self-adjoint algebras; the unital case is dense).

A complex function algebra is self-adjoint when it contains each pointwise conjugate, unital when it contains all constants, and point-separating when it distinguishes every distinct pair (Self-adjoint complex function algebras, unitality, and point separation).

Under $\mathbb{C}={\mathbb{R}}^2$, $d_{\mathbb{C}}(z,w)=\mathrm{\mid }z-w\mathrm{\mid }$ is the Euclidean metric (The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane).

For $z=a+bi$, $\bar{z}=a-bi$ and $\mathrm{\mid }z\mathrm{\mid }=\sqrt{a^2+b^2}$ (Real and imaginary parts, complex conjugation, and modulus).

Complex conjugation is a real-field automorphism with $\overline{z+w}=\bar{z}+\bar{w}$, $\overline{zw}=\bar{z}\bar{w}$ and $\overline{\stackrel{‾}{z}}=z$; and for every $z,w\in \mathbb{C}$, $z\bar{z}=\mathrm{\mid }z{\mathrm{\mid }}^2$, $\mathrm{\mid }zw\mathrm{\mid }=\mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }w\mathrm{\mid }$ and $\mathrm{\mid }z+w\mathrm{\mid }\le \mathrm{\mid }z\mathrm{\mid }+\mathrm{\mid }w\mathrm{\mid }$ (Conjugation is an involutive real-field automorphism, $z\bar{z}=\mathrm{\mid }z{\mathrm{\mid }}^2$, and modulus is definite, multiplicative, and subadditive).

The complex numbers form a field containing $\mathbb{R}$ ($\mathbb{C}=\mathbb{R}[x]\mathrm{/}(x^2+1)$ is a field, every element is uniquely $a+bi$, and every nonzero element has inverse $(a-bi)\mathrm{/}(a^2+b^2)$).

Natural powers satisfy $z^0=1$ and $z^{n+1}=z^{n}z$, while negative integer powers of nonzero $z$ are powers of its inverse (Integer powers in the complex field).

For $n\ge 1$, a subset of Euclidean ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded (Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, clause 2).

A compact subset of a metric space is compact as a topological subspace of its metric topology, and conversely (For a metric space with its metric topology, compactness in the topological sense is compactness in the metric sense, and the two notions of compact subset coincide, clause 2).

Every metric space is Hausdorff: distinct points are separated by disjoint open balls (Distinct points of a metric space have disjoint balls around them).