Re: Simple Topology Question

On 30 Dec 2005 02:16:51 -0800, "MimsyBoro" <eytan@xxxxxxxxxxxxxxx>
wrote:

>I've seen people use this without proving it and supposedly this is
>very simple but I don't know how to prove it.
>
>Prove that every open subspace of a complete metric space is of Baire's
>2nd categrory.

Say S is a (nonempty) open set. Choose x in S, r > 0 so that
B(x,r) is contained in S, and let C be the closure of
B(x,r/2). Seems to me that if S were a countable union of
nowhere-dense sets then C would be a countable union of
nowhere-dense sets as well, which can't happen since C
is a complete metric space.

You might want to verify the details carefully; there
may be a subtle point or two. (For example, the first
thing I was going to say was let C be the closed ball
about x of radius r/2, ie {y : d(y,x) <= r/2}. I don't
quite see how that would work; it seems important
that the C I give above is the closure of an open
set. Because then if A is a closed subset of S with
empty interior it follows that the intersection of A
with C has empty interior (relative to C).)

>TIA,


************************

David C. Ullrich
.