Let $X$ be a Hausdorff space. Suppose that $C(X)$ (or $C_0(X)$) is a finitely generated $C^*$-algebra. What we can say about $X$ ? For example can we characterize its inductive dimension, axioms of countability etc. ?
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The maximal ideal space $\Delta$ of a finitely generated Banach algebra is homeomorphic to a compact subset of $\mathbb{C}^n$. On the other hand, evaluation at each point of $X$ is clearly a complex homomorphism of $C(X)$. We conclude that $X$ is homeomorphic to a subset of a compact subset of $\mathbb{C}^n$.
Edit: actually, if $X$ is not compact $C(X)$ is not a Banach algebra at all (locally compact for $C_0(X)$). In locally compact case, as Nik Weaver showed in comments, $\Delta \cup \{0\}$ is compact in $\mathbb{C}^n$, so $\Delta$ is homeomorphic to a closed subset of some Euclidean space.
Also, Stone-Weierstrass theorem shows that all such $X$ satisfy the condition.
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2$\begingroup$ To tighten this up, assuming $X$ is locally compact the condition is that it should be homeomorphic to a closed subset of $\mathbb{R}^n$ for some $n$. Its spectrum including $0$ is homeomorphic to a compact subset of $\mathbb{C}^n$, and after removing $0$ it has the form I stated. $\endgroup$ Commented Mar 13, 2016 at 14:23
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$\begingroup$ In response to the edit, compact subset of $\mathbb{R}^n$ if $C(X)$, closed subset of $\mathbb{R}^n$ if $C_0(X)$. $\endgroup$ Commented Mar 13, 2016 at 15:58
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$\begingroup$ (It is usual to acknowledge comments when editing a post.) $\endgroup$ Commented Mar 13, 2016 at 15:58
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$\begingroup$ @NikWeaver But is that all such sets? What if $X$ is not locally compact? $\endgroup$ Commented Mar 13, 2016 at 16:27
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$\begingroup$ Wait! In that case $C(X)$ will not be a Banach algebra at all! $\endgroup$ Commented Mar 13, 2016 at 16:34