Let $X$ be a compact Hausdorff space and $C(X)$ be the real valued continuous functions on $X$. Suppose that $\mathcal A$ is a (topologically) closed subalgebra of $C(X)$. Is it true that if $\mathcal A$ separates points and vanishes nowhere then $1 \in \mathcal A$? If so, could someone show me how to prove it.
The reason I ask is because for the Stone-Weierstrass theorem, we always assume that $\mathcal A$ separates points, but I've seen different sources which either assume (a) that $1 \in \mathcal A$ or (b) that $\mathcal A$ vanishes nowhere. Clearly (a) $\implies$ (b). I'm wondering if it is actually an equivalence.
Yes. I'm citing Folland's Real Analysis, second edition, theorem 4.45:
If $\mathcal A$ is a closed subalgebra of $C(X; \mathbb R)$ that separates points, then either $\mathcal A =\left\{f\in C(X; \mathbb{R})\ |\ f(x_0)=0\right\}$ for an $x_0\in X$ or $\mathcal A=C(X; \mathbb{R})$.