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I've seen this link: When is the vector space of continuous functions on a compact Hausdorff space finite dimensional?

$X$ is metric and compact. But I don't want to use the Tietz theorem. So how can I prove that if $C(X)$ (the set of all continuous functions from $X\to \mathbb{C}$) is finite dimensional then $X$ is finite?

Mina
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    Can you check whether or not the definition of compactness you are using includes separated (i.e. that the space satisfies "Hausdorff's axiom", a.k.a. Hausdorff space)? Some times people define compactness as Hausdorff's axiom + quasicompact (myself included). This implies compact and Hausdorff. (A space is "quasicompact" if it satisfies the "Borel-Lebesgue's axiom" that every open cover has a finite subcover.) – William M. Jan 25 '22 at 17:46
  • @WilliamM. No in the question it just said that $X$ is metric compact set. – Mina Jan 25 '22 at 17:47
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    Metric implies the Hausdorff axiom. – William M. Jan 25 '22 at 17:48
  • @WilliamM. If I don't know the Tietze theorem, you should I prove this? – Mina Jan 25 '22 at 18:08
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    @Mina Is your space $X$ a metric space? If so, please update your question to indicate this. – Ben Grossmann Jan 25 '22 at 18:28

4 Answers4

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Let $X$ be infinite with the trivial topology $\{\emptyset , X\}$. Then $X$ is compact, $C(X)$ consists of constant functions but $X$ is not finite

Aitor Iribar Lopez
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  • This question has been edited and now is asking a different thing. Originally it asked how to prove that a compact, not necessarily Hausdorff space with finite dimensional $C(X)$ must be finite, so my answer is a counterexample – Aitor Iribar Lopez Jan 25 '22 at 20:00
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Suppose $X$ is infinite and let $\{x_n:n\ge1\}\subset X$ be distinct points. By Urysohn's lemma we can find continuous functions $h_n:X\to[0,1]$ such that $h_n(x_{n+1})=1$ and $h_n(x_{i})=0$ for all $i=1,\dots,n$. Note that if $n<m$ then $n+1<m+1$, so $$\|h_n-h_m\|_\infty\ge|(h_n-h_m)(x_{n+1})|=|h_n(x_{n+1})-h_m(x_{n+1})|=|1-0|=1\;\;\;\;\;(\star)$$ Also, $\{h_n\}_{n=1}^\infty$ is sequence in the closed unit ball of $C(X)$ since $\|h_n\|_\infty\le1$. But $(\star)$ shows us that $\{h_n\}_{n=1}^\infty$ does not contain any Cauchy subsequence. This proves that the closed unit ball of $C(X)$ is not compact, and therefore $C(X)$ is not finite dimensional.

Just dropped in
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This question already gives much of the answer: when $X$ is finite, certainly $C(X)$ is finite-dimensional, since $C(X)$ is just $\mathbb{R}^X$.

Let the metric on $X$ be $d$.

When $X$ is infinite, we can choose a sequence of finite increasing subsets $\{S_j\}_j$ where $S_j\subseteq X$ and $|S_j|=j$. We have the natural projection $\pi_j\in C(X)\to C(S_j)$ given by "evaluation at $S_j$". Also, for each $s\in S_j$, the function $x\mapsto d(s,x)$ is a function that is zero on precisely $\{s\}$. Taking products, we can construct functions that are zero on precisely all-but-one-point of $S_j$, and then sum to get any function on $S_j$. Thus $\pi_j$ is always surjective. Moreover, define the function $g_j(x)=\prod_{s\in S_j}{d(s,x)}$; we have $\pi_j(g_j)=0$.

Suppose $C(X)$ has dimension $J$. Then $C(S_J)$ also has dimension $J$, and so $\pi_J$ is injective. Thus $g_J=0$. But since $X$ is infinite, there exists $y\in X\setminus S_J$. By the definition of a metric, we can prove each factor in $g$ is nonzero, though, and so $g(y)>0$. $\rightarrow\leftarrow$.

Jacob Manaker
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Here is a paraphrasal of JustDroppedIn's argument: if $C(X)$ is finite-dimensional, then so is its dual $C(X)^\ast$; which space can be identified as the space of real measures on $X$. The map $\delta_\bullet:X\to C(X)^\ast$ that takes a point $x\in X$ and gives the Dirac measure concentrated at $x$ is a topological embedding. If $x_1,...,x_d\in X$ are distinct, then $\delta_{x_1},....,\delta_{x_d}\in C(X)^\ast$ are linearly independent by the Urysohn Lemma.

Alp Uzman
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