Counter example to Stone Weierstrass Theorem

Question

If we miss some conditions of Stone Weierstrass Theorem, will this theorem still hold?

I have come up with counter examples when we do not have compact metric space. But what if the function algebra does not separate points, or it vanishes somewhere?

Thanks!

Answer 1

If the algebra $A$ does not separate points on $X$ (a compact metric space), then there exist distinct points $a, b\in X$ such that $f(a)=f(b)$ for all $f\in A.$ If $A$ were dense in $C(X),$ then every $f\in C(X)$ would satisfy $f(a)=f(b).$ That's a contradiction, since $f(x) = d(x,a)$ belongs to $C(X).$

If the algebra $A$ vanishes somewhere, then the constant function $1$ is not the uniform limit of functions in $A.$