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Topic: Proximity Spaces, Axioms
Replies: 6   Last Post: Jun 25, 2013 10:50 PM

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William Elliot

Posts: 1,240
Registered: 1/8/12
Re: Proximity Spaces, Axioms
Posted: Jun 25, 2013 10:50 PM
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On Tue, 25 Jun 2013, dullrich@sprynet.com wrote:
> >On Mon, 24 Jun 2013, Zeit Geist wrote:
> >> On Monday, June 24, 2013 9:11:11 PM UTC-7, William Elliot wrote:
> >> >
> >> > > > n is a proximity for S when n is a binary relation for S
> >> > > > and for all A,B,C subset S
> >> > > > AnB iff BnA (symmetry)
> >> > > > A n B\/C iff AnB or AnC (additivity)
> >> > > > AnA iff A nonnul (reflexivity)
> >> > > > Easy to prove theorems are:
> >> > > > if AnB and B subset C, then AnC;
> >> > > > if A /\ B not empty, then AnB;
> >> > > > if A is not empty, then for all B, AnB or AnS\B.
> >> > > > In addition, if AnB, then A or B is not empty.

> >> >
> >> > > > Can you from the axioms above, prove
> >> > > > AnB implies both A and B are not empty?

> >
> >I've found that you can't. Let S = {a} and
> >n = { (nulset,S), (S,S) }

>
> Almost right.


n = { (nulset,S), (S,nulset), (S,S) }

> >> > > Suppose AnB and B is empty.
> >> > > Now choose a set such that ~ ( AnC ).

> >> >
> >> > Prove such a set exists. If it does, then since by
> >> > the first theorem, for all C, AnC, you have an
> >> > immediate contradiction.

> >
> >> I really can't see a way, unless we assume that there
> >> is a set A for which there is a set C where not AnC.

> >
> >If you want to disprove A n empty set, then you need
> >some C with not AnC.
> >

> >> Now, for most topological spaces this is true.
> >> And to be honest, I'm having a hard time thinking
> >> about a space where every non-empty subset
> >> is "proximal" to every other non-empty subset.

> >
> >For A,B within a topological space (S,tau), what does AnB mean?
> >

> >> I have a feeling that such a space will "collapse"
> >> down to a single point. But I'm no expert.
> >> If it does however, then the only question is
> >> if for the entire set do we have Sn{ }.
> >>
> >> I'll ask you now.
> >> Does assuming ~ ( S n {} ) imply that for all
> >> subsets of S?

> >
> >Yes. If A n empty set, then S n empty set, by the first theorem above.
> >

> >> My feeling is we must make some additional assumption like,
> >> For all A c= S, ~( A n {} ).

> >
> >For these axioms to be equivalent to the axioms of Wikipedia
> >one would replace for all A, AnA iff A not empty with
> >not empty A implies AnA and not S n empty set.
> >
> >Given a proximity n for S, then tau = { S\cl K | K subset S }
> >is a topology for S where cl K = { x | {x}nK }, provided
> >cl empty set is empty, which it is if not S n empty set.
> >
> >If n is separated, for all x,y, ({x}n{y} iff x = y ), then
> >tau is T1 and for the third axiom, just not nulset n nulset,
> >instead of the third axiom and without not S n empty set, is needed
> >for cl empty set is empty. It's claimed that for a separated proximity
> >space, the topology is Hausdorff. I don't see that. Can you prove it?

>
>




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