Re: compact sets 2,

In article <eie7fl$msp$1@xxxxxxxxxxxxxxxx>,
Daniel Mayost <mayost@xxxxxxxxx> wrote:

Please don't top-post. If you don't know what that means, follow the
links:

http://www.xs4all.nl/~hanb/documents/quotingguide.html

http://www.caliburn.nl/topposting.html

http://www.html-faq.com/etiquette/?toppost

Edited:

In article <eie67j$2of$1@xxxxxxxxxxxxxxxx>,
Daniel Mayost <mayost@xxxxxxxxx> wrote:

In article <22439754.1162514758889.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
vsgdp <nospam@xxxxxxxxx> wrote:
Is my argument rigorous enough:

Prove the intersection of an arbitrary collection of compact subset of metric space M is compact.

Let B be the intersection of all the compact subsets.

Let (p_n) be a sequence in B. Then (p_n) is a sequence in each
subset A in the intersection. But since every A compact, (p_n) has
an accumulation point, say p, in every A. So p in B. Now the book
doesn't prove it, but it gives a reference that compact set S in a
metric space is equivalent to every sequence in S having an
accumulation point in S. So by this fact, we can conclude B is
compact.

To complete this proof, you should show that, for all compact sets A,
the limit point p in A of the sequence {p_n} is the same -- which is
easy enough.

Oops, wait a second, this won't be easy because a sequence can have
several accumulation (as opposed to limit) points.

But you can go it anyway: Find an accumulation point p in one A. Then
take a subsequence that converges to that accumulation point; then
it's a limit point, and you can go from there with the idea you had.

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================

Arturo Magidin
magidin-at-member-ams-org

.

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