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




Re: Proximity Spaces, Axioms
Posted:
Jun 25, 2013 10:50 PM


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 UTC7, 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 nonempty subset > >> is "proximal" to every other nonempty 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? > >



