the dimension of continuous functions on a compact set is finite 
Possible Duplicate:
vector space of continuous functions on compact Hausdorff space 

This is a problem  am trying to solve.  Suppose the dimension of $C(X)$ is finite where $X$ is compact and Hausdorff.  Why is $X$ finite?
I was able to show that if $X$ is finite, then the dimension of $C(X)$ is finite.  I am having trouble proving the converse.
 A: Suppose $X$ is infinite. Choose $N$ distinct points $x_n \in X$. Since $X$ is Hausdorff, we can find disjoint open sets $U_n$ such that $x_n \in U_n$. Also, each $\{x_n\}$ is closed and hence compact. Urysohn's lemma shows the existence of a continuous function $f_n$ with support in $U_n$ such that $f_n(x_n) = 1$. Now consider $\phi_\alpha=\sum_{n=1}^N \alpha_n f_n = 0$. Since $\phi_\alpha(x_n) = \alpha_n$, we see that $\alpha_n = 0$, hence the $f_n$ are linearly independent.
Since $N$ was arbitrary, it follows that $\dim C(X) = \infty$.
A: copper.hat answer above works nicely (indeed, it is exactly what I was hinting at in my comment). However I would like to submit an alternative approach, based on the fact that evaluation at a point is a continuous operation on $C(X)$. 
Suppose that $C(X)$ is finite dimensional. Then the dual space $[C(X)]^\star$ is finite dimensional too. Observe that for any point $x_0\in X$ the linear functional
$$\delta_{x_0} f = f(x_0)$$
is continuous on $C(X)$, that is $\delta_{x_0}\in [C(X)]^\star$. Also, for any finite set of points $\{x_0, x_1, x_2\ldots x_m\}$ we have that 
$$\delta_{x_0}, \delta_{x_1} \ldots \delta_{x_m} \quad \text{are linearly independent.}$$
We can conclude that $X$ is finite because otherwise we could find an infinite family of linearly independent continuous linear functionals.
