# When is the vector space of continuous functions on a compact Hausdorff space finite dimensional?

I know that the vector space of all real valued continuous functions on a compact Hausdorff space can be infinite dimensional. When will it be finite dimensional? And how will I identify that vector space with $\mathbb{R}^n$ for some $n$?

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You probably want to clarify what you mean. Your first statement says the vector space is always infinite dimensional but then you ask when is it finite dimensional. In any case, if your space is finite then the vector space is finite dimensional, with the dimension equal to the number of points in your space. –  Eric O. Korman Mar 9 '11 at 14:45

## 1 Answer

For a topological space $X$, let me write $C(X,\mathbb{R})$ for the set of continuous functions $f: X \rightarrow \mathbb{R}$. Note that $C(X,\mathbb{R})$ forms a ring under pointwise addition and multiplication which contains $\mathbb{R}$ as the subring of constant functions.

Suppose $X$ is compact Hausdorff.

1) If $X = \{x_1,\ldots,x_n\}$ is finite, then since it is Hausdorff it has the discrete topology so $C(X,\mathbb{R})$ is the set of all functions from $\{x_1,\ldots,x_n\}$ to $\mathbb{R}$. In this case a natural basis is given by "$\delta$ functions'': i.e., for $1 \leq i \leq n$, let $e_i: X \rightarrow \mathbb{R}$ be such that $e_i(x_j) = \delta_{i,j}$ (i.e., $1$ if $i = j$, $0$ otherwise). Note that in this case $e_1,\ldots,e_n$ is also a family of idempotent elements giving rise to a direct product decomposition of rings $C(X,\mathbb{R}) \cong \mathbb{R}^n$.

2) If $X$ is infinite, then for every positive integer $n$ there is a subspace $X_n$ consisting of exactly $n$ points. By Step 1, the dimension of the space $C(X_n,\mathbb{R})$ is $n$. Moreover, since $X$ is Hausdorff, $X_n$ is closed in $X$, and, since $X$ is compact Hausdorff, the Tietze Extension Theorem applies to show that every continuous function on $X_n$ extends to a continuous function on $X$. In other words, the natural restriction map $r_n: C(X,\mathbb{R}) \rightarrow C(X_n,\mathbb{R})$ is surjective. Since $r_n$ is an $\mathbb{R}$-linear map (indeed a homomorphism of $\mathbb{R}$-algebras), it follows that

$\operatorname{dim} C(X,\mathbb{R}) \geq \operatorname{dim} C(X_n,\mathbb{R}) = n.$

Since $n$ was arbitrary, we conclude that $\operatorname{dim} C(X,\mathbb{R})$ is infinite.

Added: As Qiaochu points out, it is enough to require that $X$ be what I call "C-separated": i.e., for the continuous $\mathbb{R}$-valued functions to separate points of $X$. And for this it is enough that $X$ be Tychonoff. (Recall compact Hausdorff $\implies$ normal $\implies$ Tychonoff.) In fact these considerations come up in $\S 5.2$ of my commutative algebra notes. See especially the second exercise in that section, which asks for a justification of the claim I made in the first sentence of this paragraph.

Indeed, since "C-separated" implies Hausdorff, looking back at my answer it shows that one can replace "compact Hausdorff" with "C-separated" and the result still holds. However, one cannot replace "C-separated" with "regular". As I have mentioned elsewhere, there are infinite regular spaces $X$ in which the only continuous functions are the constant functions, so $C(X,\mathbb{R}) = \mathbb{R}$ is a one-dimensional space.

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I don't think you need the full power of Tietze extension; Urysohn's lemma already shows that continuous functions separate points. –  Qiaochu Yuan Mar 9 '11 at 15:22
There's also the notion of a en.wikipedia.org/wiki/Completely_Hausdorff_space — a.k.a. functionally Hausdorrff. If $X$ is completely Hausdorff then $X$ will be "C-separated" by definition. Finding a completely Hausdorff space that's not Tychonoff is left as an exercise for the reader (or you can read the Wikipedia article, I guess…) –  kahen Mar 9 '11 at 15:58
@kahen: thanks for your comment. Upon checking the wikipedia article, it seems that "completely Hausdorff" = "functionally Hausdorff" = "C-separated". Which of the former two terms is more common? Perhaps I should adjust my terminology accordingly. –  Pete L. Clark Mar 9 '11 at 16:08
I don't know. Stephen Willard uses "completely Hausdorff" in his General Topology and only parenthetically mentions "functionally Hausdorff", so I'd presume that the former is more common in the litterature. At least at the time he wrote the book –  kahen Mar 9 '11 at 16:17
Counterexamples in Topology uses "Urysohn", which makes sense to me because of the connection with the lemma, but that usually means something else. –  Chris Eagle Mar 9 '11 at 18:30