Talk:Stone-Weierstrass Theorem

I believe that $X$ does need to be Hausdorff, but that fact is implied by our other requirements. Assuming I made no mistakes, this is proven in Topological Space Separated by Mappings is Hausdorff, since $\AA$ separates the points of $X$ and $\RR$ is itself Hausdorff. --CircuitCraft (talk) 18:55, 8 October 2023 (UTC)

Good call. Added to theorem as an "also presented as" section. --prime mover (talk) 20:03, 8 October 2023 (UTC)

I see what's happened. The theorem works as an if and only if. That is, $\AA$ is dense if and only if it separates points. For the only if, you need that $X$ is Hausdorff, and there are trivial counterexamples otherwise (e.g. the indiscrete topology on $X$ where only constant maps are continuous). We probably want both directions here. Your proof is of course entirely correct. Caliburn (talk) 10:14, 9 October 2023 (UTC)

I cannot remember why I put that template, but probably, I just overlooked the fact. Obviously, I was about to apply Riesz-Kakutani Representation Theorem. Usually, one simply assumes Hausdorff in such situation, which turns out unnecessary in this special case. --Usagiop (talk) 20:38, 9 October 2023 (UTC)