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

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Peter Percival

Posts: 1,299
Registered: 10/25/10
Re: Nhood Space
Posted: Jul 1, 2013 4:02 AM
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William Elliot wrote:
> On Sun, 30 Jun 2013, William Elliot wrote:
>> On Sun, 30 Jun 2013, Peter Percival 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

>>>
>>> Is this the same as neighbourhood space defined as follows.
>>>
>>> (S, N), S a set, N a map S -> PPS (P for power set) and
>>> i) x in S => N(x) =/= 0
>>> ii) x in S, M in N(x) => x in M
>>> iii) x in S, M in N(x) => (L superset M => L in N(x)
>>> iv) x in S, L, M in N(x) => L intersect M in N(x)
>>> v) x in S, M in N(x) => exists L in N(x) s.t.
>>> L subset M and, forall y in L, L in N(y)

>
> Let's see. Define N(x) = { A | {x} << A }
> 1. For all x, {x} << S.
> 2. A << B implies A subset B
> 3. A << B and B subset C imply A << C
> 4. A << B/\C iff A << B, A << C
> 5. {x} << S and for all y in S, {y} << S. (Seems trivial.)
>
> Conversely given N(x), how is A << B to be defined?
> {x} << A when A in N(x) ?
> A << B when for all a in A, {a} << B ?
>
> Thus my second question is poignant.
> If for all a in A, {a} << B, does A << B?
>
> If we take A << B to be cl A subset int B and let
> A = { 1/n | n in N } and B = [0,1], then not A << B.
>
> So that rejects my second question leaving me wondering
> how to define A << B from the N(x)'s.
>
> I think your nhood space is more general that my proximal nhood space.
> How is an open set defined using N(x)'s?


A set is open if it is a neighbourhood of each of its points.


> U open when for all x in U, some V in N(x) with V subset U?
> Thus empty set is open.
>
> It seems forthwith, that the collection of open sets defines
> a topology and every topology gives a nhood space, which if
> derived from N(x)'s, will give the same nhood space in return.
>
> It seems proximal nhood spaces don't generate every topology
> and accordingly less general. They were, in fact, not intended
> to describe topological spaces but merely to generalize uniform
> spaces.
>



--
I think I am an Elephant,
Behind another Elephant
Behind /another/ Elephant who isn't really there....
A.A. Milne



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