points separation and dimensions Suppose $\mathcal{A}$ is a sub-algebra of $C([0,1],\mathbb{R})$. 
If $\mathcal{A}$ separates points in $[0,1]$, does it follow $\dim\mathcal{A}=\infty$?
 A: 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  $\dim\mathcal{A}<\infty$.
A: While the above answer completely answered the OP's question, there is an elementary approach, which actually gives a little bit more: Let $f$ be a non-constant, continuous function on $[0,1]$. Then the set $\{f, f^2, f^3, \ldots\}$ (here $f^n$ is the $n$th power of $f$), as a subset of $C([0,1])$, is linear independent. Indeed, if not, then $P(f)=0$ on $[0,1]$ for some non-zero polynomial $P$. This would imply that the set $f([0,1])$ is contained in the set of all roots of $P$, a finite set. This contradicts the fact that $f$ is non-constant. 
