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!

The answer is likely no, it doesn't, or else these weaker conditions would be how stone weierstrass is presented. — MT_, Mar 08 '16 at 20:45
Related question: Essential conditions for Stone-Weierstrass theorem — J.-E. Pin, Mar 08 '16 at 20:47

score 7 · Accepted Answer · answered Mar 08 '16 at 22:01

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 $A.$

Counter example to Stone Weierstrass Theorem

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