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Let $X$ be a completely regular (Tychonoff) topological space. It is known that if $\mathscr F\subseteq C(X,[0,1])$ separates points and closed sets (that is, for every closed set $E\subseteq X$ and $x\in X\setminus E$, $\exists f\in\mathscr F$ such that $f(x)\notin\operatorname{cl}[f(E)]$), then $X$ can be densely embedded into a compact Hausdorff space. Namely, considering the compact Hausdorff space $[0,1]^{\mathscr F}$ (with the product topology), there exists an embedding $e:X\to e(X)\subseteq [0,1]^{\mathscr F}$ such that the projections satisfy $\pi_{f}(e(x))=f(x)$ for all $x\in X$ and $f\in\mathscr F$. Defining $Y$ to be the closure of $e(X)$ in $[0,1]^{\mathscr F}$, $(Y,e)$ is called a Hausdorff compatification of $X$ associated to $\mathscr F$.

Now let $BC(X)$ denote the space of bounded continuous complex-valued functions on $X$. An algebra $\mathscr A\subseteq BC(X)$ (that is, a vector space that contains also the product of any two of its members) is called completely regular if (i) $\mathscr A$ is closed; (ii) $1\in\mathscr A$ ($1$ denotes the constant function); (iii) $\mathscr A\cap C(X,[0,1])$ separates points and closed sets.

It can be shown that if $(Y,e)$ is a Hausdorff compatification of $X$, then $\mathscr A_Y\equiv\{F\circ e\,|\,F\in C(Y)\}$ is a completely regular subalgebra of $BC(X)$. Moreover, if $(Y,e)$ and $(Y',e')$ are two Hausdorff compatification that give rise to the same completely regular algebra $\mathscr A_Y=\mathscr A_{Y'}$, then $(Y,e)$ and $(Y',e')$ must be homeomorphic. If one is willing to identify homeomorphic compactifications, it follows that the map $(Y,e)\mapsto \mathscr A_Y$ from Hausdorff compatifications of $X$ to completely regular subalgebras of $BC(X)$ is injective.

What I want to show is that this map is actually also surjective. That is,

Conjecture:$\quad$ Any given completely regular subalgebra of $\mathscr A\subseteq BC(X)$ is equal to $\mathscr A_Y$ for some Hausdorff compactification $(Y,e)$.

The following fact is known:

  • If $(Y,e)$ is the Hausdorff compatification of $X$ associated to some $\mathscr F\subseteq C(X,[0,1])$, then $\mathscr A_Y$ is the smallest closed subalgebra of $BC(X)$ that contains $\mathscr F$ and the constant function $1$.

Hence, given a completely regular algebra $\mathscr A\subseteq BC(X)$, in order to construct a Hausdorff compatification $(Y,e)$ for which $\mathscr A=\mathscr A_Y$, an obvious candidate would be to take $\mathscr F=\mathscr A\cap C(X,[0,1])$. What I am unable to show is that $\mathscr A$ is really the smallest closed subalgebra that contains its intersection with $C(X,[0,1])$. If this conjecture is not true, can one at least find such an $\mathscr F$ that $\mathscr A\cap C(X,[0,1])\subseteq \mathscr F\subseteq C(X,[0,1])$ and the resulting Hausdorff compactification $(Y,e)$ gives exactly $\mathscr A_Y=\mathscr A$?

Any hints and comments would be greatly appreciated.


UPDATE: Note that for any compactification $(Y,e)$, $\mathscr A_Y$ is closed under complex conjugation. Indeed, if $f\in \mathscr A_Y$, then $f=F\circ e$ for some $F\in C(Y)$. Since $\overline F\in C(Y)$, it follows that $\overline f=\overline F\circ e\in \mathscr A_Y$. Hence, in order for the conjecture above to be true, it must be the case that every completely regular subalgebra of $BC(X)$ is closed under complex conjugation. I am wondering whether this happens to be the case.

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Here's a counterexample if you don't assume your algebra is self-adjoint. Let $X$ be the unit disk in $\mathbb{C}$ endowed with the discrete topology. Let $B$ be the algebra of complex-valued functions on $X$ that extend continuously to the one-point compactification. Note that every element of $B$ is constant on a cocountable set. Let $A$ be the closed algebra generated by $B$ and the inclusion function $z:X\to\mathbb{C}$. Since $B$ is completely regular, so is $A$. I claim that $A$ is not self-adjoint, and thus cannot correspond to any compactification of $X$.

Indeed, if $A$ were self-adjoint, then $\bar{z}$ would be in $A$, and thus would be the uniform limit of a sequence of polynomials in $z$ with coefficients in $B$. For any such sequence of polynomials, all the coefficients will be constant outside of some countable set. Thus if $\bar{z}$ is in $A$, then there is a sequence of constant-coefficient polynomials in $z$ that converges uniformly to $\bar{z}$ outside a countable set. But this is clearly impossible (e.g., because powers of $z$ are orthogonal to $\bar{z}$ in $L^2$ of the disk).

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  • $\begingroup$ Great counterexample, thank you very much! $\endgroup$ – triple_sec Jul 13 '14 at 23:47
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Yes, although the terminology is unusual this is standard. The compactifications of $X$ correspond to quotients of $\beta X$ which correspond to closed subalgebras of $C(\beta X) \cong BC(X)$. For example, see Theorem 3.55 of my book Measure Theory and Functional Analysis.

If you don't care to look it up, the idea is to define an equivalence relation on $\beta X$ by setting $x \sim y$ if $f(x) = f(y)$ for all $f \in \mathcal{F}$. The key point is that this is a closed equivalence relation --- it is closed as a subset of $\beta X \times \beta X$ --- and this implies that $\beta X/\sim$ is compact Hausdorff (not as obvious as it looks and a good exercise; Theorem 1.45 of my book).

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    $\begingroup$ Another comment: In your Theorem 3.55, you assume that $\mathscr A$ is self-adjoint, i.e., it is closed under complex conjugation. However, in my context, I never assumed this. Does it somehow follow from it being completely regular (i.e., that its intersection with $\mathscr A\cap C(X,[0,1])$ separates points and closed sets)? $\endgroup$ – triple_sec Jul 13 '14 at 20:09
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    $\begingroup$ Oh, I didn't notice that you left that out. No, of course you need to assume the algebra is self-adjoint. Eric gives a counterexample in his answer. $\endgroup$ – Nik Weaver Jul 13 '14 at 23:29
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    $\begingroup$ (Incidentally, it is, remarkably, the case that norm closed ideals are automatically self-adjoint. But that is not directly relevant to your question.) $\endgroup$ – Nik Weaver Jul 13 '14 at 23:30
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    $\begingroup$ Also, Eric's example falsifies the conjecture about $\mathcal{A}$ being the smallest subalgebra that contains its intersection with $C(X,[0,1])$. Even simpler, let $X$ be the open unit disc and let $\mathcal{A}$ be the set of bounded analytic functions on $X$. Then $\mathcal{A} \cap C(X,[0,1])$ contains only constant functions. $\endgroup$ – Nik Weaver Jul 13 '14 at 23:38
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    $\begingroup$ Bounded analytic functions are not "completely regular", though; the idea behind my example was basically to enlarge them to make them completely regular. $\endgroup$ – Eric Wofsey Jul 13 '14 at 23:42

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