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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: Jul 1, 2013 10:57 AM
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On Sun, 30 Jun 2013 19:03:00 -0700, William Elliot <marsh@panix.com>
wrote:

>On Sun, 30 Jun 2013, dullrich@sprynet.com wrote:
>> 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

>>
>> (i) this is an actual definition you saw somewhere

>
>Yes, you'll find it included with proximity spaces in
>Wikipedia, Willard and other places.
>

>> _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

>
>It's a generalization of uniform spaces.
>

>> A << B implies S\B << S\A
>
>If A << B is interpeted to mean
>cl A subset int B, then cl S\B subset int S\A.


Ah, that's different.

>
>As you've shown, A << B can't be interpeted to mean A subset int B.
>

>> 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.)

>
>This indicates why the topology inheritent in nhood, or proximity spaces are
>general topological spaces. If fact, their Tychonov.
>

>> >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!

>
>Read it in the consted of nhood spaces.
>

>> 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"?
>>

>Include them as or if needed. Separated nhood spaces are T1
>and separated normal nhood spaces are Hausdorff.
>
>Often normal is included as part of the definition.
>

>> > 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.

>
>Correct.
>

>> >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...)
>>

>Nhood spaces are the DeMorgan like duals of proximity spaces.
>Williard gives a closure operator cl, for a proximity space
>and leaves it as an exercise to show cl is a closure operator.
>The part I'm having trouble with is proving cl cl A = cl A,
>which in the dual nhood space is int int A = int A.
>

>> >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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