Proving an equivalent statement for the Stone-Weierstrass theorem In my metric space course, we were taught the Stone-Weierstrass theorem as follows 

We were told however that the second condition ("A contains the constant functions") may be replaced by the condition "A contains the constant function 1". Is someone able to show me how this is possible, I cannot locate a proof of this fact. Thanks for any help in advance ! 

This is false. Consider the algebra generated by the functions $1$ and $x+1$ on $[0, 1]$. This algebra separates points because it contains the function $x+1$, yet every function $f$ in this algebra satisfies $f(x)\geq 1$ (because the generators $1$ and $x+1$ do, and this property is preserved under multiplication and addition) and thus the algebra cannot approximate the continuous function $0$.
