Give an example of a countable dense subset of $C[0,1]$.
I think the set of all rational polynomials defined on $[0,1]$ will do.
My attempt: Let $A$ be the set of all rational polynomials defined on $[0,1]$. Clearly $A\subset C[0,1]$, pick any $f\in C[0,1]$ and consider a ball $B_{\epsilon}(f)$ for $\epsilon>0$. By Stone Weierstrass theorem $\exists$ polynomial $P$ such that $P \in B_{\epsilon}(f)$, in fact we can choose rational polynomial, so $B_{\epsilon}(f) \cap A \neq \phi ,\ \forall \ \epsilon>0$. Therefore $cl(A) = C[0,1]$
Will this work?
