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

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William Elliot

Posts: 2,383
Registered: 1/8/12
Re: Nhood Space -- proof here
Posted: Jul 3, 2013 5:09 AM
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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
> Look at axiom 5. It is different from what
> is in Willard or on Wikipedia.
> 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
> 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.

> "If there exists a point x which is close to
> both A and B, then A is close to B"

{x}nA, {x}nB ==> AnB is immediate.

Just use {x}n{x} and for all A,B,C
AxB, B subset C ==> AxC
which is a quick result of additive.

> 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

How does
{x}n{y}, {y}nA ==> {x}nA
show cl cl A = cl A? I don't get it.

> To me, the use of the underlying set theory is cheating.

Hardly so, for immediately, with the definition of n as a
binary relation between subsets of S, you are tossed into
the winds of set theory.

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