Let $X$ be a compact Hausdorff space. Assume that the vector space of real-valued continuous functions on $X$ is finite-dimensional. I would like to conclude that $X$ is finite.
Certainly, the converse implication is trivial.
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For each $x \in X$, consider the linear map that sends $f$ to $f (x)$. – Zhen Lin Feb 18 '16 at 20:56
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@ZhenLin Is the map 1-1 or onto? – Dhrubajyoti Sarkar Feb 18 '16 at 20:59
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It's definitely not $1-1$. @jaggu – Thomas Andrews Feb 18 '16 at 21:02
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Is it onto?@ThomasAndrews – Dhrubajyoti Sarkar Feb 18 '16 at 21:05
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Sure, for any $r\in\mathbb R$, you can define the constant function $f_r:X\to \mathbb R$ so that $f_r(x)=r$. However, I don't think knowing if this map is "onto" is the key here. @jaggu – Thomas Andrews Feb 18 '16 at 21:07
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By contraposition, suppose that $X$ is infinite. By the Hausdorff property, one may construct recursively a countably infinite family $(O_n)_{n=1}^\infty$ of pair-wise disjoint, non-empty open subsets of $X$. For each $n$ pick $x_n\in O_n$. As $X$ is normal, by Urysohn's lemma, for each $n$ there exists a continuous function $f_n$ such that $f_n(x_n)=1$ and $$\{x\in X\colon f_n(x)\neq 0\}\subseteq O_n.$$ Clearly, the set $\{f_n\colon n\in \mathbb{N}\}$ is linearly independent, so $C(X)$ is infinite-dimensional.
Tomasz Kania
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