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?

  • $\begingroup$ What is $g$ in the definition of $A'$? $\endgroup$ – Taras Banakh Feb 4 '18 at 15:51
  • $\begingroup$ 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. $\endgroup$ – afton Feb 4 '18 at 16:34
  • $\begingroup$ Oh sorry! I should have noticed that! $\endgroup$ – Daron Feb 4 '18 at 17:09
  • $\begingroup$ $g$ is just an arbitrary element of $A$. That should be clear now. $\endgroup$ – Daron Feb 4 '18 at 19:18
  • $\begingroup$ 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. $\endgroup$ – KP Hart Sep 24 '18 at 8:41

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.