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Topic: Nhood Space
Replies: 24   Last Post: Jul 3, 2013 10:43 PM

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William Elliot

Posts: 2,217
Registered: 1/8/12
Re: Nhood Space
Posted: Jun 30, 2013 10:03 PM
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On Sun, 30 Jun 2013, dullrich@sprynet.com wrote:
> On Sat, 29 Jun 2013 21:39:59 -0700, William Elliot <marsh@panix.com>
> wrote:

> >(S,<<) is a nhood space when << is a binary relation for P(S) and
> >for all A,B,C subset S
> > empty set << A << S
> > A << B implies A subset B
> > A << B implies S\B << S\A
> > A << B/\C iff A << B and A << C

> (i) this is an actual definition you saw somewhere

Yes, you'll find it included with proximity spaces in
Wikipedia, Willard and other places.

> _If_ (ii), then it seems likely that this is an
> attempt to axiomatize the relation "A is a
> nbd of B" in a topological space. But it
> definitely does not do that. If we start
> with a topological space and then define
> << by saying A << B if B is a nbd of A
> then the axiom

It's a generalization of uniform spaces.

> A << B implies S\B << S\A

If A << B is interpeted to mean
cl A subset int B, then cl S\B subset int S\A.

As you've shown, A << B can't be interpeted to mean A subset int B.

> does not hold. (For example, if A is any
> open set then A << A. But S\A = C can
> be any closed set, and a closed set C need
> not be a nbd of itself.)

This indicates why the topology inheritent in nhood, or proximity spaces are
general topological spaces. If fact, their Tychonov.

> >A << B is taken to mean B is a nhood of A.
> >Thus {x} << A would mean A is a nhood of x.
> >
> >Additional axioms are separation
> > for all x,y, if x /= y, then {x} << S\y
> >and normality
> > for all A,B, if A << B, then there's some K with
> > A << K << B
> >
> >Useful theorems

> The statements below cannot be theorems, because
> there are no hypotheses!

Read it in the consted of nhood spaces.

> Of course it's easy to guess what you intend the hypotheses to be - the
> problem is that there are at least two natural guesses. Presumably you're
> assuming we have a Nhood space. _Are_ you assuming the "additional axioms"?

Include them as or if needed. Separated nhood spaces are T1
and separated normal nhood spaces are Hausdorff.

Often normal is included as part of the definition.

> > are
> >A << B, B subset C implies A << C
> >A subset C, B << C implies A << C

> That must be a typo for
> A subset B, B << C implies A << C.


> >Define the interior of a set A, int A = { x | {x} << A }.
> >Easy theorems are
> > int empty set = empty set; int S = S
> > int A/\B = int A /\ int B; int A subset A
> > A subset B implies int A subset int B.
> >
> >How would one prove int int A = int A?

> I don't know. I don't have a counterexample,
> but I tend to doubt that it's true. (I also tend
> to suspect I'm correct in my speculations above,
> meaning that you simply don't have the definition
> of "Nhood space" right...)

Nhood spaces are the DeMorgan like duals of proximity spaces.
Williard gives a closure operator cl, for a proximity space
and leaves it as an exercise to show cl is a closure operator.
The part I'm having trouble with is proving cl cl A = cl A,
which in the dual nhood space is int int A = int A.

> >Since int A subset A, int int A subset int A.
> >So the question actually is how to prove
> > int A subset int int A?
> >
> >Another question.
> >If for all x in A, {x} << B, is A << B provable?
> >
> >From the axiom
> > A << B/\C iff A << B and A << C
> >
> > S\B \/ S\C << S\A iff S\B << S\A and S\C << S\A
> >
> >Thus for all A,B,C
> > B \/ C << A iff B << A, C << A
> >
> >Hence if A if finite and for all x in A, {x} << B, then A << B.
> >What if A isn't finite?

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