Can we make this subspace $\aleph_0$-dimensional?

Question

Let $X$ be a compact Hausdorff space and $A\subseteq X$ a subspace of $X$. Is it possible for the space $\{f\vert_A:f\in\mathcal{C}(X,\mathbb{R})\}$ to be $\aleph_0$-dimensional as a $\mathbb{R}$-vector subspace of $\mathcal{C}(X,\mathbb{R})$? I know that in this case $\mathcal{C}(X,\mathbb{R})$ is, at least, $\mathfrak{c}$-dimensional as an infinite-dimensional Banach algebra. But I don't know if we can restrict it to make it $\aleph_0$-dimensional somehow. Thanks!

Answer 1

I claim that your space is the same as $\mathcal C(\bar A, \Bbb R)$. In particular the restriction map $f \mapsto f|_A$ from $\mathcal C(\bar A, \Bbb R)$ to your space is an isomorphism.

• The restriction map is a well-defined map to your space because if $f: \bar A \to \Bbb R$ is continuous then it extends to $X$ by the Tietze extension theorem.
• It's clearly surjective, as the restriction of a continuous function on $X$ is the restriction of the restriction of that function to $\bar A$.
• It's injective, because two continuous functions to $\Bbb R$ that agree on $A$ must agree on $\bar A$.

It follows that your space cannot be $\aleph_0$-dimensional, for the same reason $\mathcal C(X, \Bbb R)$ cannot be.