Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: compact
Replies: 9   Last Post: May 18, 2013 9:56 PM

 Search Thread: Advanced Search

 Messages: [ Previous | Next ]
 William Elliot Posts: 2,637 Registered: 1/8/12
Re: compact
Posted: May 15, 2013 3:06 AM
 Plain Text Reply

On Tue, 14 May 2013, Butch Malahide wrote:
> On May 14, 11:15 pm, William Elliot <ma...@panix.com> wrote:

> > A point x is an accumulation point of A
> > when for all open U nhood x, U /\ A is infinite.
> >
> > If S is compact, then every infinite set A has an accumulation point.
> > Proof.
> > If not, then for all x, there's some open U_x nhood x with finite U /\ A.
> > Since C = { U_x | x in S } covers S,
> >         there's a finite subcover { U_x1,.. U_xj }
> > Thus A = A /\ (U_x1 \/..\/ U_xj } = (U_x1 /\ A) \/..\/ (U_xj /\ A),
> > is finite, which of course it isn't.
> >
> > If every infinite A subset S has an accumulation point, is S compact?

>
> omega_1

A point x is a saturation point of A
when for all open U nhood x, |U /\ A| = |A|

If S is compact, then every infinite set A has saturation point.
Proof.
If not, then for all x, some open U_x nhood x with |U_x /\ A| < |A|.
Since C = { U_x | x in S } covers S,
. . there's a finite subcover { U_x1,.. U_xj }
Thus A = A /\ (U_x1 \/..\/ U_xj } = (U_x1 /\ A) \/..\/ (U_xj /\ A),
is finite which of course it isn't.

If every infinite A subset S has a saturation point, is S compact?

Date Subject Author
5/15/13 William Elliot
5/15/13 Butch Malahide
5/15/13 William Elliot
5/15/13 Butch Malahide
5/18/13 William Elliot
5/15/13 David C. Ullrich
5/15/13 quasi
5/16/13 David C. Ullrich
5/16/13 quasi
5/15/13 William Elliot

© The Math Forum at NCTM 1994-2018. All Rights Reserved.