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

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David C. Ullrich

Posts: 3,016
Registered: 12/13/04
Re: Nhood Space
Posted: Jun 30, 2013 10:45 AM
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On Sat, 29 Jun 2013 21:39:59 -0700, William Elliot <marsh@panix.com>
wrote:

>(S,<<) is a nhood space when << is a binary relation for P(S) and
>for all A,B,C subset S
> empty set << A << S
> A << B implies A subset B
> A << B implies S\B << S\A
> A << B/\C iff A << B and A << C


You need to clarify the context! We've been through
this before. Tell us whether

(i) this is an actual definition you saw somewhere

or

(ii) it's a definition you made up.

This makes a big difference. If (i) then I say hmm,
I don't quite get it. If (ii) then I'd say that this
definition doesn't do what you hope it does,
thus answering what is presumably your real,
unstated, question.

_If_ (ii), then it seems likely that this is an
attempt to axiomatize the relation "A is a
nbd of B" in a topological space. But it
definitely does not do that. If we start
with a topological space and then define
<< by saying A << B if B is a nbd of A
then the axiom

A << B implies S\B << S\A

does not hold. (For example, if A is any
open set then A << A. But S\A = C can
be any closed set, and a closed set C need
not be a nbd of itself.)

>A << B is taken to mean B is a nhood of A.
>Thus {x} << A would mean A is a nhood of x.
>
>Additional axioms are separation
> for all x,y, if x /= y, then {x} << S\y
>and normality
> for all A,B, if A << B, then there's some K with
> A << K << B
>
>Useful theorems


The statements below cannot be theorems, because
there are no hypotheses! Of course it's easy to
guess what you intend the hypotheses to be -
the problem is that there are at least two
natural guesses. Presumably you're assuming
we have a Nhood space. _Are_ you assuming
the "additional axioms"?

> are
>A << B, B subset C implies A << C
>A subset C, B << C implies A << C


That must be a typo for

A subset B, B << C implies A << C.

>Define the interior of a set A, int A = { x | {x} << A }.
>Easy theorems are
> int empty set = empty set; int S = S
> int A/\B = int A /\ int B; int A subset A
> A subset B implies int A subset int B.
>
>How would one prove int int A = int A?


I don't know. I don't have a counterexample,
but I tend to doubt that it's true. (I also tend
to suspect I'm correct in my speculations above,
meaning that you simply don't have the definition
of "Nhood space" right...)

>Since int A subset A, int int A subset int A.
>So the question actually is how to prove
> int A subset int int A?
>
>Another question.
>If for all x in A, {x} << B, is A << B provable?
>
>From the axiom
> A << B/\C iff A << B and A << C
>
> S\B \/ S\C << S\A iff S\B << S\A and S\C << S\A
>
>Thus for all A,B,C
> B \/ C << A iff B << A, C << A
>
>Hence if A if finite and for all x in A, {x} << B, then A << B.
>What if A isn't finite?
>
>
>





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