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muppet

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Thanks in advance.

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- Thread starter muppet
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muppet

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Thanks in advance.

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muppet

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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?

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mathwonk

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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.

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mathwonk

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muppet

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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...

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mathwonk

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