Re: Closing the Intersection


William Elliot wrote:
Hence you want property p for which for all A,B
p(A,B) iff cl A/\B = cl A /\ cl B ?

yes that's what I want.

Then let p(A,B) be cl A/\B = cl A /\ cl B.
or perferable take p(A,B) as
cl A /\ cl B subset cl A/\B
Otherwise consider cases
A subset B
A,B disjoint
other
For example,
A subset B or B subset A ==> p(A,B)
A,B closed ==> p(A,B)
p(A,S\A) ==> A clopen

S? .. do you mean our topological space X?
yes they are trivial facts, the last one is in fact an equivalence

p(A,X\A) <=> A clopen

A,B completely separated, ie disjoint closures ==> p(A,B)
Hm, p(A,B) iff p(B,A); p(A,A).


yes, natural observations. :) p(A,B) and p(B,C) does not imply p(A,C)
for instance in R, A=(-2,0) B=(-1,1) C=(0,2).


Perhaps you'd content yourself with property p weaker than closed
for which
p(A), p(B) ==> cl A/\B = cl A /\ cl B.


Nah, I guess this would be less useful. I can imagine that there are
lots of p's satisfying that above.

Have you a clarification of characterization?
Does
cl A /\ cl B subset cl A/\B
suffice?

Sure it does, because the other way is trivial.
Find p such that
p(A,B) iff cl A /\ cl B subset cl(A/\B)

Lets take the following case, let A, B be connected and
if A neither subset of B nor the other way round and A/\B is nonempty
then if there is a closure point of A/\B not in the closure point of A
AND closure point of B then this point must be in wlog cl(A). Now what
forces this to be a contradiction? In the euclidean topology this leads
to a contradiction. But I cannot say so for any topology. Problem is
that the euclidean topology is just too cute.

Sincerely,
Jose Capco

.

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