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

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William Elliot

Posts: 1,606
Registered: 1/8/12
Re: Nhood Space, correction
Posted: Jul 1, 2013 4:50 AM
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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?
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.





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