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.




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**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.



Are Hausdorff compactifications of a Tychonoff space $X$ in one-to-one correspondence with completely regular subalgebras of $BC(X)$?
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).


