MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


If we take a localy compact space $X$ and we put $A=C_{b}(X)$ the $\mathbb{R}$-algebra of bounded continous functions on $X$, we have an embeding of topological space

$$\psi:X\longrightarrow \text{Spec}_{\max}(A) $$ defined by $\psi(x)=I_{x}$, where $I_{x}=\{ f\in A / ~ f(x)=0 \}$, and where $\text{Spec}_{\max}(A)$ is the set of maximal ideal of the ring $A$ with the Zariski topology. Then we define the Stone-Čech compactification of $X$ by $\displaystyle \overline{X}^{sc}=\overline{\psi(X)}$.

We can also define the Alexandroff compactification of $X$ with the same method, we just must to take $A=C_{0}(X)$ the algebra of continous functions with zero limit at infinity.

My question is: Can we define all compactification of $X$ with the same method, in other hand if $\widetilde{X}$ is a compactification of $X$ ($X$ locally compact), then there is a sub-algebra $A$ of the algebra $C_{b}(X)$ such that $\widetilde{X}$ is homeomorphic with $\psi(X)$, where $$\psi:X\longrightarrow \text{Spec}_{\max}(A) $$ is defined by $\psi(x)=I_{x}$ the maximal ideal of elements $f\in A$ such that $f(x)=0$.

share|cite|improve this question
I removed the new tag named '884' as I have no idea what it should mean; if ever this had a meaning (as opposed to being created inadvertently): sorry! – quid Nov 12 '12 at 17:21
@quid: Apparently MATH 884 is the code for the Topology course at the University of New Hampshire and 52-884 is the code at Southwestern University. – Ramiro de la Vega Nov 13 '12 at 18:38

Yes. This is called Constantinescu-Cornea theorem, Ideale Ränder Riemannscher Flächen, Springer 1963. They deal with 2-dimensional manifolds only, but the result is true for any locally compact space. It is stated and proved in this generality in the book of Marcel Brelot, On topologies and boundaries in potential theory, Springer 1971.

share|cite|improve this answer

Note : In order to define the alexandrov compactification of $X$ you have to take the algebra of function converging at infinity (not just those converging to $0$)

If by a compactification of $X$ you mean a compact Hausdorff topological space $Y$ equiped a continuous map $ X \rightarrow Y$ having a dense image, then the answer is true :

If you consider instead the algebra of complex function then it is immediate from the gelfand duality :

A such space is just a compact Hausdorff topological space equipped with a surjective continuous map form $\overline{X}^{sc}$. But the gelfand duality assert that compact Hausdorff topological space corresponds to commutative $C^ * $ algebra. And it is easy to see, that a continuous map between compact space is surjective if and only if the induce map between the $C^ * $ algebra is injective (and hence isometric by classical result on $C^ * $ algebra).

So, compactification of $X$ corresponds to sub-$C^*$-algebra of $C^b(X)$. ie sub-algebra of $C^b(X)$ which are closed under complexe conjugaition, closed for the normt topology, and which contain $1$.

One can then easily see that such algebra also corresponds to sub-algebra of the algbera of real bounded function, which contain $1$ and are closed for the norm topology.

NB : local compactness is actually useless for all this to work.

share|cite|improve this answer
The compactification needs to be Hausdorff for this to work. I think you can construct non-Hausdorff compactifactions - for instance by taking the Stone-Cech compactification and gluing together all the points at infinity less one. Or do those not count? – Will Sawin Nov 12 '12 at 18:03
Yes you're right. I meant compact hausdorf everywhere. I'm just used to the French/Bourbakist convention to call "compact" a compact Hausdorff space, and quasi-compact a space which is compact but possibly non Hausdorff. – Simon Henry Nov 12 '12 at 18:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.