If $C(X)$ is finite dimensional then $X$ is finite. I've seen this link: When is the vector space of continuous functions on a compact Hausdorff space finite dimensional?
$X$ is metric and compact. But I don't want to use the Tietz theorem. So how can I prove that if $C(X)$ (the set of all continuous functions from $X\to \mathbb{C}$) is finite dimensional then $X$ is finite?
 A: Let $X$ be infinite with the trivial topology $\{\emptyset , X\}$. Then $X$ is compact, $C(X)$ consists of constant functions but $X$ is not finite
A: Suppose $X$ is infinite and let $\{x_n:n\ge1\}\subset X$ be distinct points. By Urysohn's lemma we can find continuous functions $h_n:X\to[0,1]$ such that $h_n(x_{n+1})=1$ and $h_n(x_{i})=0$ for all $i=1,\dots,n$. Note that if $n<m$ then $n+1<m+1$, so
$$\|h_n-h_m\|_\infty\ge|(h_n-h_m)(x_{n+1})|=|h_n(x_{n+1})-h_m(x_{n+1})|=|1-0|=1\;\;\;\;\;(\star)$$
Also, $\{h_n\}_{n=1}^\infty$ is sequence in the closed unit ball of $C(X)$ since $\|h_n\|_\infty\le1$. But $(\star)$ shows us that $\{h_n\}_{n=1}^\infty$ does not contain any Cauchy subsequence. This proves that the closed unit ball of $C(X)$ is not compact, and therefore $C(X)$ is not finite dimensional.
A: This question already gives much of the answer: when $X$ is finite, certainly $C(X)$ is finite-dimensional, since $C(X)$ is just $\mathbb{R}^X$.
Let the metric on $X$ be $d$.
When $X$ is infinite, we can choose a sequence of finite increasing subsets $\{S_j\}_j$ where $S_j\subseteq X$ and $|S_j|=j$.  We have the natural projection $\pi_j\in C(X)\to C(S_j)$ given by "evaluation at $S_j$".  Also, for each $s\in S_j$, the function $x\mapsto d(s,x)$ is a function that is zero on precisely $\{s\}$.  Taking products, we can construct functions that are zero on precisely all-but-one-point of $S_j$, and then sum to get any function on $S_j$.  Thus $\pi_j$ is always surjective.  Moreover, define the function $g_j(x)=\prod_{s\in S_j}{d(s,x)}$; we have $\pi_j(g_j)=0$.
Suppose $C(X)$ has dimension $J$.  Then $C(S_J)$ also has dimension $J$, and so $\pi_J$ is injective.  Thus $g_J=0$.  But since $X$ is infinite, there exists $y\in X\setminus S_J$.  By the definition of a metric, we can prove each factor in $g$ is nonzero, though, and so $g(y)>0$.  $\rightarrow\leftarrow$.
A: Here is a paraphrasal of JustDroppedIn's argument: if $C(X)$ is finite-dimensional, then so is its dual $C(X)^\ast$; which space can be identified as the space of real measures on $X$. The map $\delta_\bullet:X\to C(X)^\ast$ that takes a point $x\in X$ and gives the Dirac measure concentrated at $x$ is a topological embedding. If $x_1,...,x_d\in X$ are distinct, then $\delta_{x_1},....,\delta_{x_d}\in C(X)^\ast$ are linearly independent by the Urysohn Lemma.
