In the question ([$C(X)$ as finitely generated $C^\*$-algebra](https://mathoverflow.net/questions/233512/cx-as-finitely-generated-c-algebra)), the answer show that spectrum of an abelian unital finitely generated C\*-algebra is homeomorphic to compact subset of $\mathbb{C}^{n}$. I would like to know more about the details of the proof of this fact. Also I would like to know, if the number of generators is related to the $n$ in anyway?


Thank you in advance.



finitely generated C*-algebra as $C(X)$
Let $X$ be a compact Hausdorff space. Let $C(X)$ be the set of all continuous functions $f:X\rightarrow\mathbb{C}$. I claim that $X$ embeds into $\mathbb{C}^{n}$ if and only if the $C^{\*}$-algebra $C(X)$ is generated by $n$ functions. To prove this fact, we need to the complex Stone Weierstrass theorem.


We say that a subset $\mathcal{A}\subseteq C(X)$ separates points if whenever $x,y\in X,x\neq y$, there is some $f\in\mathcal{A}$ with $f(x)\neq f(y)$.


Theorem: (Complex Stone Weierstrass theorem) Suppose that $\mathcal{A}$ is a closed dense subalgebra of $C(X)$ (here $C(X)$ is given the topology induced by the supremum metric) that separates points and where if $f\in\mathcal{A}$, then $\overline{f}\in\mathcal{A}$. Then $\mathcal{A}=C(X)$.


Suppose that $C(X)$ is generated as a $C^{\*}$-algebra by $f\_{1},\dots,f\_{n}:X\rightarrow\mathbb{C}$. Then whenever $x,y\in X,x\neq y$, there is some $i$ with $f\_{i}(x)\neq f\_{i}(y)$. Therefore, the mapping
$f\_{1}\times\dots\times f\_{n}:X\rightarrow\mathbb{C}^{n}$ is continuous and injective, and since $X$ is compact, the mapping $f\_{1}\times\dots\times f\_{n}$ is an embedding.


Suppose to the contrary that $f\_{1}\times\dots\times f\_{n}:X\rightarrow\mathbb{C}^{n}$ is an embedding. Then whenever $x,y\in X$, there is some $i$ with $f\_{i}(x)\neq f\_{i}(y)$, so $f\_{1},\dots,f\_{n}$ satisfies the conditions of the complex Stone Weierstrass theorem. We conclude that $f\_{1},\dots,f\_{n}$ generates the $C^{\*}$-algebra $C(X)$.


More generally, if $\kappa$ is a cardinal, then $C(X)$ is generated by at most $\kappa$ many functions if and only if $X$ embeds into $\mathbb{C}^{I}$ for some set $I$ with $|I|\leq\kappa$, but the the smallest infinite $|I|$ where $X$ embeds into $\mathbb{C}^{I}$ is actually a well-known cardinal invariant.


Proposition: Suppose $X$ is an infinite compact Hausdorff space. Then the least cardinal $\kappa$ such that $C(X)$ is generated by at most $\kappa$ many functions is precisely $w(X)$ ($w(X)$ is the size of the smallest basis of $X$).


