
Re: Nhood Space
Posted:
Jun 30, 2013 10:45 AM


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? > > >

