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## Topology Q+A Board

From: Henno Brandsma
Date: Nov 20, 2005
Subject: Re: X has a metrizable compactification iff X is 2nd countable and tychonoff

In reply to "X has a metrizable compactification iff X is 2nd countable and tychonoff", posted by jacob on Nov 20, 2005:
>I need to prove:
>X has a metrizable compactification iff X is 2nd countable and Tychonoff.
>
>
>It looks like we will need to use Urysohn's Metrization Theorem (among
>other things), but I haven't been able to come up with the details.
>
>Any help would be greatly appreciated.

Yes. Let X be second countable and Tychonov. Then we can embed X into [0,1]^N
by the way the Tychonov theorem is usually proved: for every pair (U, V), where U,V are from
the countable base (that exists for R) and satisfying cl(U) subset V, we pick a Tychonov function
(it is a lemma / well-known fact that a second countable Tychonov space is normal)
f = f(U,V): X --> [0,1] such that f == 0 on cl(U) and 1 on X\V.
Then enumerate these pairs and functions as (U_n, V_n) and f_n,
we send x in X to ( f_n(x) ) (n in N) in [0,1]^N, and show this is an embedding.

So X embeds into the compact metric space [0,1]^N and so its closure in that space
is a metrisable compactification of X.

The other way round: if X is dense in a compact metrisable space Y, then Y
is second countable and Tychonov (a compact metric space is second countable
and a compact Hausdorff space is normal, hence Tychonov) and so are all its
subspaces, including X.

Henno