vitali sets

[raghav...@yahoo.com]

The intersection of a vitali set which is also bernstein with unit
interval has outer measure 1. How do we show that given a number x
with |x| < = 1, there is a vitali set whose intersection with unit
interval is x?(Is the statement true?)
Any answer or reference would be appreciated.

[raghav...@yahoo.com]

[The World Wide Wade]

In article
<3e231197-743e-4b99...@i24g2000prf.googlegroups.com>,
raghav...@yahoo.com wrote:

Please give appropriate definitions.

[madhur...@gmail.com]

>
> Please give appropriate definitions.

vitali set - consider the equivalence classes with equivalence
relation x - y = rational.(on real line). Assuming AC,there is a set
which contains exactly one element from each equivalence class. That
is the vitali set.

Bernstein set - A set which intersects with every perfect set but
contains no perfect set is called a bernstein set.

Both happen to be non-measurable.

[Dave L. Renfro]

raghav_ty...@yahoo.com wrote:

Did you try googling sci.math?

See the following sci.math thread:

Outer measure of Vitali sets
http://tinyurl.com/42fwy3

For more about Bernstein sets, see:

Essay on Bernstein sets
http://groups.google.com/group/sci.math/msg/36cdfd0c767b1ab3
http://groups.google.com/group/sci.math/msg/64a01ce3d199a641

For an interesting construction due to Sierpinski that doesn't
seem to be very well known, see:

http://groups.google.com/group/sci.math/msg/5f95ae8bad97a9f8
http://matwbn.icm.edu.pl/ksiazki/fm/fm31/fm3111.pdf

Dave L. Renfro

[raghav...@yahoo.com]

Yes, I did google sci.math. Thanks for the links anyway. They don't
seem to answer my question.
Given a number x < 1, does there exist a vitali set such that its
intersection with any unit interval(or maybe even any measurable set
of measure 1) has outer measure x.

[Robert Israel]

raghav...@yahoo.com writes:

... with, of course, x > 0.

Yes to the first version, no to the second (even if you replace measurable
with closed).

1) For each integer n, let F_n be the family of open subsets of [n, n+x]
with measure < x, and let F = {{t}: t in R} union union_n F_n. This has
cardinality c, so well-order it in such a way that every member of F has fewer
than c predecessors. Since [n,n+x] \ A has cardinality c for any A in F_n,
there is a function f: F -> R such that
(a) for each A in F_n, f(A) is in [n,n+x] \A and is not in {f(B)+q: B a
predecessor of A, q in Q}, and
(b) for each t in R, if f(B) - t is in Q for some predecessor B of {t},
then f({t}) = f(B) for the first such B. Otherwise, f({t}) is some member
of (t + Q) intersect [0,x].

Then S = { f(A): A in F } is a Vitali set, i.e. contains exactly one
representative of each coset, S is contained in union_n [n,n+x],
and S intersect [n,n+x] has outer measure x. For any real a, if
n = floor(a) and a-n >= x then S intersect [a,a+1] = S intersect [n+1,n+1+x]
while if a-n < x then
S intersect [a,a+1] = (S intersect [a,n+x]) union (S intersect [n+1,a+1])
which in each case has outer measure x.

2) If S_n = S intersect [n,n+1] has outer measure x, it is contained in
an open set U_n of measure < (x+1)/2, and C_n = [n,n+1]\U_n is a closed
set of measure > (1-x)/2 disjoint from S. Take the union of these for
enough n, and you get a closed set of measure > 1 whose intersection
with S is empty.
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada