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From the book "Putnam and Beyond."

The problem:

Show that the interval [0, 1] cannot be partitioned into two disjoint sets A and B such that B = A + a for some real number a.

Proof:

Assume that A,B, and a satisfy A∪B =[0,1], A∩B =∅, 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. An inductive argument 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.

I don't really understand what's going on:

An inductive argument shows that for any positive integer n, the interval (1−(2n+1)a, 1−2na] is in B.

How did he come to this conclusion?

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 wording here is very confusing. Is he merely pointing to the fact that [0, 1] is a closed interval, while the other is open? — "This interval exists [0, 1]" is not a sentence that makes any syntactical sense to me.

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    The verb that was used was *exits* not *exists*. It simply means that as $n$ increases, the small interval drifts to the left, and eventually leaves (=exits) the interval $[0,1]$. Just before the exit, it must contain $0$.2011-06-23
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    The induction really is straight forward: $(1-a,1]$ is contained in $B$, so $(1-2a,1-a]$ is contained in $A$, so $(1-3a,1-2a]$ is contained in $B$, so $(1-4a,1-3a]$ is contained in $A$,...2011-06-23

2 Answers 2