Stone-Weierstraß theorem

Let $𝑌$ be a compact space and $𝑋$ be a ∗-subalgebra of the continuous function ∗-algebra $𝖳 𝗈 𝗉 (𝑌, ℂ)$ that is separating. Then $𝑋$ is dense in $𝖳 𝗈 𝗉 (𝑌, ℂ)$ , fun i.e. $C l (𝑋) = 𝖳 𝗈 𝗉 (𝑌, ℂ)$ .

Proof

proof

Finite version

Let $𝑋 \leq 𝕂 𝑌$ be a subalgebra of the function algebra $𝕂 𝑌$ on a finite set $𝑌$ . Then $𝑋 = 𝕂 𝑌$ iff $𝑋$ separates points, linalg i.e. for any distinct $𝑥, 𝑦 \in 𝑌$ there exists $𝑓 𝑥, 𝑦 \in 𝑋$ so that $𝑓 𝑥, 𝑦 (𝑥) = 1$ and $𝑓 𝑥, 𝑦 (𝑦) = 0$

Proof

Assume $𝑋$ is separating. For each $𝑥 \in 𝑌$ , let $𝛿 𝑥 \in 𝕂 𝑌$ so that $𝛿 𝑥 (𝑦) = [𝑥 = 𝑦]$ ^[Invoking an Iverson bracket.]. Defining $𝑓 𝑥, 𝑦$ as above, it follows
$𝛿 𝑥 = \prod 𝑦 \neq 𝑥 𝑓 𝑥, 𝑦$
and since ${𝛿 𝑥} 𝑥 \in 𝑌$ span $𝕂 𝑌$ it follows $𝑋 = 𝕂 𝑌$ . For the converse just set $𝑓 𝑥, 𝑦 = 𝛿 𝑥$ for all $𝑦 \neq 𝑥$ .

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Stone-Weierstraß theorem

Finite version

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