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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 5:09 AM


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.



