Suppose $\mathcal{A}$ is a sub-algebra of $C([0,1],\mathbb{R})$. 


If $\mathcal{A}$ separates points in $[0,1]$, does it follow $\dim\mathcal{A}=\infty$?



points separation and dimensions
Without loss of generality, we may assume that $\mathcal{A}$ contains the constant functions (otherwise, just add them to $\mathcal{A}$, it will increase the dimension by $1$). Hence, $\mathcal{A}$ satisfies the conditions of Stone-Weierstrass theorem, and so is dense in $\mathcal{C}([0,1],\mathbb{R})$.


Now if $\dim\mathcal{A}<\infty$, it is close in $\mathcal{C}([0,1],\mathbb{R})$ (every finitely dimensional subspace of a normed space is closed), and since it is also dense, we get that $\mathcal{A}=\mathcal{C}([0,1],\mathbb{R})$, which clearly contradicts to $\dim\mathcal{A}<\infty$.


