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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.

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john
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  • I submit a trick that you might find useful. Suppose that $f_1\ldots f_m \in C(X)$ have pairwise disjoint supports. Then they are linearly independent, because if you take a vanishing linear combination $$\lambda_1f_1+\ldots+\lambda_mf_m=0$$ and evaluate it at points in the support of $f_j$ you get $\lambda_j=0$. So, suppose by contradiction that $\dim C(X)<\infty$ and that $X$ is infinite. Can you find an infinite sequence $f_1, f_2\ldots f_m \ldots $ in $C(X)$ with pairwise disjoint supports? If you can then you are finished. – Giuseppe Negro Dec 24 '12 at 18:11

2 Answers2

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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 $\dim C(X) = \infty$.

copper.hat
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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.

Giuseppe Negro
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