Re: Analysis with nonmeasurable set..
- From: magidin@xxxxxxxxxxxxxxxxx (Arturo Magidin)
- Date: Tue, 22 Jan 2008 15:51:19 +0000 (UTC)
In article <fn51k5$fhn$1@xxxxxxxxxxxxxxxx>,
mina_world <mina_world@xxxxxxxxxxx> wrote:
Hello sir~
My quesion is simple, but previous content is complex. sorry.
There exists a nonmeasurable set in the interval [0,1).
pf)
Def)
If x and y are real numbers in [0,1),
x +' y = x + y, x + y < 1.
= x + y - 1, x + y >= 1.
Def)
If E is a subset of [0,1),
E +' y = {z | z = x +' y, x in E}
We define an equivalence relation '~' in the set I = [0,1)
by saying that x and y in I are equivalent,
to be written x~y, if x~y is rational.
Clearly, the relation ~ partitions the set I into mutually disjoint
equivalence classes;
that is, any two elements of the same class differ by a rational number
while those of the different classes differ by an irrational number.
Construct a set P by choosing exactly one element from each equivalence
class.
(by the axiom of choice.)
Clearly P subset [0,1).
We shall now show that P is a nonmeasurable set.
Let {r_i} be an enumeration of rational numbers in [0,1) with r_0 = 0.
Define P_i = P +' r_i. P_0 = P.
(a) P_m /\ P_n = empty if m =/= n.
(b) Union_n P_n = [0,1)
Proof of (a)
Let, if possible, y in P_m /\ P_n.
Then there exist p_m and p_n in P such that
y = p_m +' r_m = p_n +' r_n.
But p_m - p_n is a rational number, whence p_m ~ p_n.
Since P has only one element from each equivalence class,
we must have m = n.
This is a contradiction.
so, If m =/= n, then P_m /\ P_n = empty.
Namely, <P_i> is pairwise disjoint set-sequence.
Proof of (b)
Let x in [0,1).
Then x lies in one of the equivalent classes and
as such x is equivalent to an element y of P.
Suppose r_i is the rational number by which x differs from y.
Then x in P_j for some j (*****)
and hence [0,1) subset Union_n P_n
while the reverse inclusion is obviously true.
so, Union_n P_n = [0,1).
[...]
----------------------------------------------------------
I can't understand (*****) part.
Namely,
Suppose r_i is the rational number by which x differs from y.
Then x in P_j for some j.
I think...
The relation ~ partitions the set I into mutually disjoint equivalence
classes.
Suppose that {A, B, C, ....} is disjoint equivalence classes.
If x in A, there y in P such that |x - y| in Q(rational).
Namely, {x, y} subset A.
If x - y = r_i in Q /\ [0,1), x = y + r_i.
Since x in [0, 1), x = y + r_i in [0,1).
so, x = y + r_i in P + r_i.
so, x in P +' r_i.
so, x in P_i.
If y - x = r_i in Q /\ [0,1), [Namely, x - y < 0]
How do you show that x in P_j for some j ?
That is, what do you do if y>x? Then look at x+1. (x+1)-y=r_i in
Q/\[0,1), you have that x+1 = y+r_i; since x+1 is in [1,2), what you
have is that x = y +' r_i, so x is in P +' r_i, hence x is in P_i.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org
.
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