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$X$ be a compact Hausdorff space such that $\dim$ $C(X,\mathbb{R})<\infty$ we need to show $|X|<\infty$, I must say I have no idea how to prove this result. please help. Thank you!

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Can you show that you can always extend a function defined on a finite set of points on $X$ to a continuous function on all of $X$? –  Zhen Lin Feb 7 '13 at 11:51
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See also: math.stackexchange.com/q/250325/49437 for a generalization. –  Martin Feb 7 '13 at 15:04

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up vote 4 down vote accepted

Let $N$ be the dimension of $C(X,\Bbb R)$. Assume that there are $N+1$ distinct points $x_1,\dots,x_{N+1}$ in $X$. As $X$ is compact and Hausdorff it is normal, hence by the Urysohn lemma, we can find for each $j$ a continuous function $g_j\colon X\to\Bbb R$ such that $g_j(x_k)=0$ when $k\neq j$ and $g_j(x_j)=1$.

The family $\{g_j,1\leqslant j\leqslant N+1\}$ is necessarily linearly dependent, so we can assume that $g_{N+1}=\sum_{j=1}^Ng_j$. Evaluating this equality at $x_{N+1}$ yields a contradiction.

This proves that $\dim C(X,\Bbb R)=\operatorname{card}(X)$.

The result is not necessarily true when $X$ is not assumed to be Hausdorff. For example, take $X$ an infinite set with the topology $\{\emptyset,X\}$. The only continuous functions are constant ones.

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Perhaps it's worth pointing out where we used the compactness assumption... –  Zhen Lin Feb 7 '13 at 13:12
    
@ZhenLin: Even I can't see where compactness is used. –  S.C. May 30 at 8:39
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You need some assumption on $X$ to apply Urysohn's lemma. –  Zhen Lin May 30 at 9:47
    
You are right. This is indeed where compactness is used (I should have done this edit one year ago...). –  Davide Giraudo May 30 at 9:51

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