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




Re: Nhood Space  proof here
Posted:
Jul 3, 2013 10:43 PM


On Tue, 2 Jul 2013, fom wrote:
> http://planetmath.org/sites/default/files/texpdf/39037.pdf > at bottom of first page > Great find!
> Look at axiom 5. It is different from what > is in Willard or on Wikipedia.
Equivalent form of normality.
> However, it matches the axioms on page 93 of > > http://books.google.com/books?id=FdThSVDYTrgC&pg=PA106&lpg=PA106&dq=%22local+proximity+space%22&source=bl&ots=NkGa0_oXAE&sig=CZkYlfEUX0X2x99ZR1xkeTVD3Fk&hl=en&sa=X&ei=VpbTUe7NHIS7ywHnrIHABA&ved=0CDEQ6AEwAQ#v=onepage&q=%22local%20proximity%20space%22&f=false
Not possible to view because I've imediately "reached the viewing limit".
Now that you've given proof of cl cl A = cl A, I see the roll normality plays in the proof and use that to give a simple and direct proof int int A = int A = { x  {x} << A } without any mind twisting double negatives. But first noted that A << B implies A subset int B.
Since int A subset A, int int A subset int A. To prove int A subset int int A, assume x in int A. Thus {x} << A and by normality there's some K with {x} << K, K << A. Hence by the note above; {x} subset int K, K subset int A. Consequently x in int K subset int int A, QED.
To continue with the insprition you've proveded note in passing that int S\A = S\cl A. Thus the following follows from A << B: S\B << S\A; S\B subset int S\A = S\cl A; cl A subset B
Using that result and normality with A << B, there's some K with A << K << B. Whence cl A subset K subset int B.
In conclusion, A << B implies cl A subset int B (in fact some K with cl A subset int K, cl K subset int B) within the topology { int A  A subset S } of the proximity nhood space (S,<<).
That however doesn't imply normal pnhood spaces induce normal topologies, as tempting it is to think it would.
> I believe you will find that there are > some uses of underlying set theory that > one cannot justify on the basis of the > axioms alone. > > Based on the axioms alone, the problem I found > is that > > {y} n cl(A) <> ( ( {y} n A ) \/ ( {y} n cl(A)/A ) ) > > may be true, but > > ~( {y} n A ) /\ ( {y} n cl(A)/A ) > > cannot be shown to be problematic. Perhaps > I am just too stupid. In defense of myself, > this reminds me of the sequence of derived > sets associated with Cantor (and, I believe, > descriptive set theory). > > In the second link, on page 93 you will find > the remark: > > "If there exists a point x which is close to > both A and B, then A is close to B" > > Once again, I think this is from the underlying > set theory. But, it also makes the proof > trivial. > > That is, if > > ( {y} n {x} ) /\ ( {x} n A ) > > then > > ( {y} n A ) > > (uses symmetry axiom) > > and cl cl A subset cl A > > ================================ > > To me, the use of the underlying set > theory is cheating. When I first read > about proximity spaces, I took them to > characterize vagueness. Using the > underlying set theory in the proofs > seems like an assumption of separability > regardless of how the nearness relation > itself may be defined. > > Oh well, I am wrong about many things. > > > > > > > > > > >



