Re: intersection of compact sets

From: Michael Barr (barr_at_barrs.org)
Date: 03/28/05 

Date: 27 Mar 2005 16:33:11 -0800

Lawrence House <lawrence.house@comcast.net> wrote in message news:<17474488.1111894965271.JavaMail.jakarta@nitrogen.mathforum.org>...
> This seems complicated to me. In general a compact set is one that
has the property that any open covering of it can be reduced to a
finite subcovering. In Rn however it is simpler-i.e. a compact set is
any set which is closed and bounded. A theorem says this. Furthermore
an arbitrary intersection of closed sets is closed and so an arbitrary
intersection of compact sets is compact. ( I think this is true in any
topological space not just Rn.) So from this we have that the
intersection of F(a) for every a is a closed subset of the open set G.
But I don't know where to go from here. I'll keep trying.

Here is how I see it. Assume each F(a) is compact closed. Also, we
can assume wlog that for some a_0, every F(a) \subseteq F(a_0) (pick
a_0 arbitrarily and repace F(a) by F(a) \cap F(a_0).) Now let K(a) =
F(a) - G. Then {K_a} is a collection of closed subsets of a compact
space with empty intersection, which implies that a finite subset of
them already has empty intersection. The intersection of the
corresponding subset of the F(a) must lie in G.

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