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

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Peter Percival

Posts: 951
Registered: 10/25/10
Re: Nhood Space
Posted: Jul 2, 2013 3:03 AM
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William Elliot wrote:
> On Mon, 1 Jul 2013, Peter Percival wrote:
>> Peter Percival wrote:
>>> William Elliot 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

>>>
>>> Is this the same as neighbourhood space defined as follows.
>>>
>>> (S, N), S a set, N a map S -> PPS (P for power set) and
>>> i) x in S => N(x) =/= 0
>>> ii) x in S, M in N(x) => x in M
>>> iii) x in S, M in N(x) => (L superset M => L in N(x)
>>> iv) x in S, L, M in N(x) => L intersect M in N(x)
>>> v) x in S, M in N(x) => exists L in N(x) s.t.
>>> L subset M and, forall y in L, L in N(y)
>>> ?

>>
>> What I meant when asking are they the same was:
>> a) can << be defined in terms of N, and can William's axioms be deduced as
>> theorems from mine; and

>
> LIkely not for the reasons I previously indicated.
>

>> b) can N be defined in terms of <<, and can my axioms be deduced as theorems
>> from William's?

>
> Yes, I showed you how.
>

>> And of course everyone knew that's what I meant.
>
> Popular Presumption.
>

>> I wrote of my axioms just for reasons of euphony, I think they are
>> Hausdorff's. Again, someone more knowledgeable than I will know.

>
> Here's from the web. Are these equivalent to yours?
>
> The axioms formulated by Hausdorff (1919) for his concept of a
> [32]topological space. These axioms describe the properties satisfied
> by subsets of elements x in a [33]neighborhood [34]set E of x .
>
> 1. There corresponds to each point x at least one [35]neighborhood
> U(x) , and each [36]neighborhood U(x) contains the point x .
>
> 2. If U(x) and V(x) are two [37]neighborhoods of the same point x ,
> there must exist a [38]neighborhood W(x) that is a subset of both.
>
> 3. If the point y lies in U(x) , there must exist a [39]neighborhood
> U(y) that is a [40]subset of U(x) .
>
> 4. For two different points x and y , there are two corresponding
> [41]neighborhoods U(x) and U(y) with no points in common.


Mine doesn't imply the separation axiom 4.


--
I think I am an Elephant,
Behind another Elephant
Behind /another/ Elephant who isn't really there....
A.A. Milne



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