Re: Simple question about dense linear orders

On Fri, 30 May 2008 19:18:41 EDT, HanGookIn wrote:
Let [A,<] be a **countable** dense linear order without endpoints.
Let A' be a subset in A such that A' is dense (everywhere) in A.

Then A\A' is discreet?
If yes, could I see a proof? Thank you.

For a counterexample, take A = Q with the natural order and let A' be the dyadic
rationals (those whose denominators are powers of 2).

--
Dave Seaman
Third Circuit ignores precedent in Mumia Abu-Jamal ruling.
<http://www.indybay.org/newsitems/2008/03/29/18489281.php>
.