Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

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


Math Forum » Discussions » sci.math.* » sci.math.independent

Topic: Open and Shut
Replies: 10   Last Post: Feb 4, 2013 10:50 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Virgil

Posts: 7,021
Registered: 1/6/11
Re: Open and Shut
Posted: Feb 4, 2013 3:08 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

In article
<aef313bf-935a-47e5-a5f9-1781ba694ce8@h17g2000yqe.googlegroups.com>,
Butch Malahide <fred.galvin@gmail.com> wrote:

> On Feb 3, 11:34 pm, Virgil <vir...@ligriv.com> wrote:
> > In article <Pine.NEB.4.64.1302031827150.2...@panix1.panix.com>,
> >  William Elliot <ma...@panix.com> wrote:
> >
> >
> >
> >
> >

> > > On Sun, 3 Feb 2013, Virgil wrote:
> > > >  William Elliot <ma...@panix.com> wrote:
> >
> > > > > A subset A, of an ordered set is convex when
> > > > > for all x,y in A, for all a, (x <= a <= y implies a in A).

> >
> > > > > I will call an interval an order convex subset of Q.
> > > > > Given an interval, what's the probablity that it's
> > > > >  open, closed, both, neither?

> >
> > > > The only probability that is certain in Q is that the probability of
> > > > being both open and closed is zero, as Q and {} are the only non-empty
> > > > order-convex sets in Q that are both open and closed under the order
> > > > toology, and there are infinitely many other intervals which are not
> > > > both open and closed.

> >
> > > (-pi,pi) /\ Q is a proper, not empty, clopen, order convex subset of Q.
> >
> > Depends on which topology one has for Q.

>
> There is only one "natural" topology for Q. The order topology of Q
> coincides with the subspace topology of Q as a subspace of R. This is
> not true for every subset of R (e.g. consider [0,1) union [2,3)); it
> is true for Q because Q is a dense subset of R.
>

> > If one uses the order topology on Q, in which a basis of the interiors
> > of intervals with ENDPOINTS IN Q, Not in R, then your set is not closed
> > in Q.

>
> Wrong. It is true that the open intervals of Q with rational endpoints
> constitute a base for the topology of Q (a vector space has a BASIS, a
> topology has a BASE), and the set (-pi, pi) /\ Q is not an element of
> that base, but it is an open set in Q because it is a union of
> elements of the base. This is analogous to the fact that, in the more
> familiar setting of the real line, the collection of all intervals
> with rational endpoints is a base for the topology, while an open
> interval with one or both endpoints irrational is also an open set,
> being a union of open intervals with rational endpoints.


Oops! My error! I stand corrected!
--





Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.