Can we express separability of a ray-remainder in terms of the function algebra?

Let $X = [0, 1)$ be a ray and $C(X)$ the algebra of bounded continuous real functions. The spectrum of $C(X)$ is the Stone-Cech compactification $\beta [0,1)$ of the ray. It's easy to see the compactification is separable. It's somewhat harder to see the remainder $\beta [0,1) - [0,1)$ is not separable.

Suppose we consider instead the algebra $C(X)$ of functions on $X$ that tend to some limit. The spectrum of this algebra is just the arc $[0,1]$ and the remainder is a singleton which is separable.

More generally suppose $A \subset C(X)$ is an arbitrary closed unital subalgebra. For convenience replace $A$ with $A' = \{f \in C(X):$ there exists $x \in [0,1)$ and $g \in A$ such that $f$ and $g$ coincide on $[x,1)\}$. This ensures $A'$ separates points of [0,1) and the spectrum contains a copy of the ray. Again we have a remainder and by Gelfand duality it is a continuous image of the Stone-Cech remainder.

Is there a nice property of the algebras $A$ or $A'$ that is equivalent to the associated remainder being separable? By nice I mean it does not come down to expressing properties of the remainder in terms of maximal ideals and hull-kernels?

• What is $g$ in the definition of $A'$? – Taras Banakh Feb 4 '18 at 15:51
• There is a fatal flaw in the formulation---in order to get the Stone-Čech compactification as spectrum you need to consider the Banach algebra $C^b(X)$ of bounded continuous functions. The space $X$ is locally compact so that if we supply $C(X)$ with its natural topology (compact convergence), its spectrum is just $X$. I think you have to restate your problem to give it some content. – afton Feb 4 '18 at 16:34
• Oh sorry! I should have noticed that! – Daron Feb 4 '18 at 17:09
• $g$ is just an arbitrary element of $A$. That should be clear now. – Daron Feb 4 '18 at 19:18
• To get an idea of what you are looking for consider Boolean algebras and their Stone spaces: the Stone space of a Boolean algebra is separable iff the algebra itself is $\sigma$-centered (the union of countably many families with the finite intersection property). There is no getting around using subsets of the ring in the characterization. – KP Hart Sep 24 '18 at 8:41