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Existence of x,y so that x-y is in Z-{0}.

  1. Jan 11, 2015 #1

    WWGD

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    Hi, let E be a measurable subset of the Real line with m(E)>1 . I want to show
    there are x,y in E so that x-y is in ## \mathbb Z-{0} ##. My idea is to restrict the
    quotient ## \mathbb R / \mathbb Z |_E ##. This quotient cannot be contained in
    [0,1], since m([0,1])=1 and m(E)>1. From this I want to show that there must be
    more than one representative of each clase of the quotient in E , but I am having
    trouble tightening up the argument. Any ideas?

    Thanks.
     
  2. jcsd
  3. Jan 12, 2015 #2

    jbunniii

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    Suppose there are no such ##x,y \in E##.

    For each integer ##k##, let ##E_k = E \cap [k, k+1)##. Then the ##E_k##'s are measurable and pairwise disjoint, with ##\sum m(E_k) = m(E)##.

    Now consider the translated sets ##F_k = E_k - k##, in other words, ##F_k## is ##E_k## translated by ##k##, so ##F_k \subset [0,1)##.

    If ##F_k \cap F_m \neq \emptyset## for some ##k \neq m##, then there are ##x_k \in E_k## and ##x_m \in E_m## such that ##x_k - k = x_m - m##, so ##x_k - x_m \in \mathbb{Z} \setminus \{0\}##, contrary to our assumption above. So the ##F_k##'s are pairwise disjoint.

    Now Lebesgue measure is translation-invariant, so each ##F_k## is measurable with ##m(F_k) = m(E_k)##. By countable additivity,
    $$m\left(\bigcup_{k \in \mathbb{Z}} F_k\right) = \sum_{k \in \mathbb{Z}} m(F_k) =\sum_{k \in \mathbb{Z}} m(E_k) = m(E) > 1$$
    But on the other hand, each ##F_k \subset [0,1)## for each ##k##, and therefore also ##\bigcup_{k \in \mathbb{Z}} F_k \subset [0,1)##, which means that
    $$m\left(\bigcup_{k \in \mathbb{Z}} F_k\right) \leq 1$$
    So we have a contradiction.
     
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