# Minimal compactification of an infinite dimensional space

## Main Question or Discussion Point

Wikipedia seems fairly consistent in stating that infinite-dimensional topological vector spaces such as Hilbert space aren't locally compact, which means that they can't have a one-point compactification. As metric spaces they're Tychonoff spaces, and thus can be compactified with the Stone-Cech compactification, but this is the "maximal" construction. Does anyone know of a minimal compactification of such manifolds, in the sense that it obtains the smallest possible compact extension of such a space?

## Answers and Replies

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Here's a procedure I just came up with, which seems to suggest that I could compactify the space with at most one extra point

In what follows any "facts" I quote will probably derive from wikipedia articles; I'd appreciate anything contentious being drawn to my attention.

Let our infinite-dim vector space, assumed metrizable, be called X. As a metric space it's compactly generated, and hence we can infer the existence of a locally compact hausdorff space Y such that X is the quotient space of Y under some map. As Y is a locally compact hausdorff space, it admits a one-point compactification. Then apply the original quotient relation to obtain X', our original point set with at most one extra point if the point at infinity should prove inequivalent to members of X. As the quotient space of a compact space, X' is compact.

Is there a flaw in this procedure? As it seems that only locally compact Hausdorff spaces admit one-point compactifications, if this does result in a compact space it seems that it must do some great violence to the original topology; would there be a way of showing whether or not properties such as being hausdorff were preserved by this procedure?

mathwonk
Homework Helper
Its sort of obvious that Hilbert space is not locally compact since one can presumably find an infinite orthogonal sequence of unit vectors.

I don’t know the answer to this interesting question, but seem to recall (from a class 45+ years ago) a relevant fact. An inclusion from a completely regular T1 space X into a compact such space Y, induces by restriction an injection from the algebra of continuous functions on Y to a uniformly closed subalgebra of bounded continuous point separating functions on X containing the constants. Conversely any such subalgebra of BC(X) recovers the compactification Y. The largest compactification Y is the one associated to the full algebra BC(X), and a smallest compactification would come from a smallest such subalgebra, if one exists. When X is locally compact, one can consider the subalgebra of continuous functions on X having “limits at infinity”, i.e. such that there exists L such that for every e>0, |f-L| < e, everywhere off some compact set.
Then the closure of the embedding of X in the Tychonoff cube defined by these functions gives the one point compacitification.

Just the mumblings of an old man with a kid’s memory from math 212.

mathwonk
Homework Helper
you might want to ask this at stack exchange, mathematics section, where lots of mathematicians answer these questions.

Thanks for your replies, mathwonk- as you might have noticed from my posts in other subforums here, I'm a theoretical physicist without much of a brain for pure maths. I'm glad you found it interesting anyway!

If you figure out an answer, let us know. It's an incredibly interesting problem!!

The closest anwer I can give is to consider the projective space associated with the infinite dimensional vector space. I have a feeling that this is a rather small compactification. But I didn't check the details yet, so I don't even know if it's a compactification at all...

mathwonk
I was thinking about my attempt yesterday whilst bored in a seminar. If the "facts" I quoted from wikipedia are indeed facts, then it looks as if my argument constructs a one-point compactification of the original set X; define the equivalence relation on the one-point compactification of the locally compact set Y by the union of the equivalence relation on Y that leads to X with $$\{(\infty,\infty)\}$$; then the equivalence classes form X along with a single addition. I got stuck when it came to thinking about the topology on whatever set it is that has elements of an infinite dimensional hilbert space as equivalence classes, but (particularly as a result of mathwonk's post) I'm inclined to say that the end result can't be hausdorff, whatever it is.