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

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Peter Percival

Posts: 2,623
Registered: 10/25/10
Re: Nhood Space, correction
Posted: Jul 1, 2013 5:13 AM
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William Elliot wrote:
> On Mon, 1 Jul 2013, Peter Percival wrote:
>> William Elliot 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.)

> This will not do. Define int U = { x | {x} << U }.
> Now again is the need to prove int int U = int U.
> With than,
> if {x} << U, {x} << int U subset U,
> for all y in int U, {y} << int U
> will prove the fifth axiom
> If {x} << U: x in int U = int int U; {x} << int U
> If y in int U: y in int int U; {y} << int U
> Any notions how to prove int int A = A?

>>> 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.

> That's simpler and equivalent to what I surmised below.
> Simularly an open set in a p-nhood space would be a set U with
> for all x in U, {x} << U. The showing int int U = int U would
> be equivalent to showing int U is open.

>>> U open when for all x in U, some V in N(x) with V subset U?
>>> Thus empty set is open.

> Hoe would the interior of a set A, be defined?

x is in the interior of A if A is a neighbourhood of x.

> As the largest open set contined in A?


>>> 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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