Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


Math Forum
»
Discussions
»
sci.math.*
»
sci.math
Notice: We are no longer accepting new posts, but the forums will continue to be readable.
Topic:
Bounded almost everywhere
Replies:
5
Last Post:
Jun 17, 2011 7:55 AM



fishfry
Posts:
1,368
Registered:
12/6/04


Re: Bounded almost everywhere
Posted:
Jun 16, 2011 12:18 PM


In article <20110615233422.U18415@agora.rdrop.com>, William Elliot <marsh@rdrop.remove.com> wrote:
> Let S = QxQ as a subset of R^2. > > A rational ray is a ray from the origin O, with rational slope. > An irrational ray is a ray from O with rational slope. > > Along each rational ray, S is unbounded > and along each irrational ray, S is bounded. > > The rational rays are almost nowhere. > The irrational rays are almost everywhere. > > S is unbounded almost nowhere, bounded almost everywhere.
I don't entirely follow your point that this takes place in QxQ. A line through the origin with irrational slope can NOT pass through ANY point of QxQ, say (q1, q2), because such a point has rational slope q2/q1.
So your irrational rays lie entirely in the complement of QxQ.
In other words no point of QxQ lies on any irrational ray.
Can you try to clarify what you mean by calling an irrational ray unbounded?



