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

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 William Elliot Posts: 2,601 Registered: 1/8/12
Nhood Space
Posted: Jun 30, 2013 12:39 AM

(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

A << B is taken to mean B is a nhood of A.
Thus {x} << A would mean A is a nhood of x.

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

Date Subject Author
6/30/13 William Elliot
6/30/13 David C. Ullrich
6/30/13 William Elliot
7/1/13 David C. Ullrich
7/1/13 David C. Ullrich
7/1/13 fom
7/2/13 William Elliot
7/2/13 Peter Percival
7/2/13 William Elliot
7/2/13 fom
7/3/13 fom
7/3/13 William Elliot
7/3/13 William Elliot
6/30/13 fom
6/30/13 Peter Percival
6/30/13 fom
7/1/13 Peter Percival
7/1/13 William Elliot
7/1/13 Peter Percival
7/1/13 William Elliot
7/1/13 Peter Percival
7/1/13 Peter Percival
7/2/13 William Elliot
7/2/13 Peter Percival
7/2/13 William Elliot