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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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    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.2012-12-24

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