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$\hspace{0.44cm}$Given a compact Hausdorff topological space $(X, \tau)$($X$ is an infinite set and $\tau$ non-trivial), and a proper subalgebra $\mathfrak{A}$ in $C(X)$, we call $\mathfrak{A}$ an SW-Algebra iff $\mathfrak{A}$ separates points, contains constant functions and self-adjoint. If $\mathfrak{A}_0$ is a proper subalgebra in $C(S)$ that only separates points and self-adjoint, then $\{c + f\,\vert\,c \in \mathbb{C}, f \in \mathfrak{A}_0\}$ will be an SW-Algebra. Also, it would be easy to create a self-adjoint algebra based on a a given algebra $\mathfrak{B}$. Then the main difficulty will be finding a separating set in $C(X)$ and then we can create an algebra based on that set.

$\hspace{0.44cm}$According to this post, the existence of such a set implies $X$ is metririzable. If $X$ is metririzable then $C(X)$ will be separable and such a set will exist. Meanwhile, by Urysohn's Lemma any countable set in $X$ can be separated by a countable set of continuous functions but how about the converse? If any countable set of continuous functions cannot separate $X$, does a subset in $X$ have to be countable to be separated by countably many continuous functions?

Sanae
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  • Of course not, for the last question; $X=[0,1]$ has a countable separating set and is not countable. – Henno Brandsma Feb 04 '21 at 07:05
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    You cannot have something weaker: all such conditions imply $X$ is metrisable (and so is $C(X)$). – Henno Brandsma Feb 04 '21 at 07:07
  • Thanks for the comment. It is true that "there exists a countable separating set of continuous functions in $C(X)$" iff "$X$ metririzable" but I am simply curious if there is something that is not equivalent to "$X$ metririzable". Also I need to edit my last question. – Sanae Feb 04 '21 at 13:34
  • If you formulate some condition C that implies that there is a countable separating family, C implies that $X$ is metrisable by that equivalence. So C is at least as strong as that. So yould say that C is $X$ is a closed connected subset of $[0,1]^\omega$ which is strictly stronger... Trivial example, granted.. – Henno Brandsma Feb 04 '21 at 13:55
  • I see. This is clear now and I will only keep my last question. – Sanae Feb 04 '21 at 14:04
  • I think you get a counterexample by taking $X$ to be the disjoint union of your favorite uncountable compact metric space, say $[0,1]$, with your favorite non-metrizable compact Hausdorff space, say $\omega_1 + 1$. The single function $f$ with $f(x)=x$ on $[0,1]$ and $f(x)=0$ on $\omega_1+1$ separates $[0,1]$. – Nate Eldredge Feb 04 '21 at 14:14
  • In other words, this seems to be asking "Can a non-metrizable compact Hausdorff space have an uncountable metrizable subset?" and the answer is clearly yes. – Nate Eldredge Feb 04 '21 at 14:16
  • @NateEldredge That is a nice example. It would be also interesting to know if I can always find a copy of $[0,1]$, i.e. the most well-known compact metric space, in any non-metririzable compact Hausdorff space. But I will leave this to myself. – Sanae Feb 04 '21 at 14:24
  • You certainly can't find a copy of $[0,1]$ in $\omega_1 + 1$ because it's totally disconnected. Moreover, I think you could show that every metrizable subset of $\omega_1 + 1$ is countable. – Nate Eldredge Feb 04 '21 at 14:28
  • @NateEldredge My bad .... I shouldn't have said $[0, 1]$ but "a copy of compact metric space". Again $\omega_1 + 1$ is another nice counterexample. – Sanae Feb 04 '21 at 14:47
  • "Banach spaces of continuous functions" by Semadeni is a good reference on $C(X)$ spaces for $X$ compact. Lots of classic facts are shown there. – Henno Brandsma Feb 04 '21 at 15:11
  • @HennoBrandsma I am sorry for missing your comment yesterday. This is indeed a nice book to read and thank you again for your answers. – Sanae Feb 05 '21 at 19:20

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