1Assume thatA, B, and a satisfyA∪B = [0, 1], A∩B = âˆ
, B = A+a.
We can assume
that a is positive; otherwise, we can exchange A and B.
Then (1 − a, 1] ⊂ B; hence
(1 − 2a, 1 − a] ⊂ A.
by induction we can show that shows that for any positive integer n,
the
interval (1−(2n+1)a, 1−2na] is in B, while the interval (1−(2n+2)a, 1−(2n+1)a] is in A.
However, at some point this sequence of intervals leaves [0, 1].
The interval of
the form (1−na, 1−(n−1)a] that contains 0 must be contained entirely in either A or B, which is impossible since this interval exits [0, 1].
The contradiction shows that the
assumption is wrong,
and hence the partition does not exist.
