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

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William Elliot

Posts: 1,434
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.


Correct.

> >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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