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

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William Elliot

Posts: 1,610
Registered: 1/8/12
Re: Nhood Space -- proof here
Posted: Jul 3, 2013 10:43 PM
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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 p-nhood 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.
>
>
>
>
>
>
>
>
>
>
>




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