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All but finitely many theorem

  1. Mar 27, 2012 #1
    "All but finitely many" theorem

    Let [itex]S[/itex] be a set with cardinality [itex]|S|=\aleph_0[/itex]. Let [itex]A,B \subseteq S[/itex]. Let [itex]S\backslash A[/itex] and [itex]S\backslash B[/itex] be finite. Then [itex]A \cap B \neq \varnothing[/itex].

    How can this be shown? I came across it as an assumption in a proof that a sequence in a metric space can converge to at most one limit.
     
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  3. Mar 27, 2012 #2

    Fredrik

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    Re: "All but finitely many" theorem

    There's something wrong with that assumption. I'll show you a counterexample.

    S = the set of integers
    A = all integers except 0
    B = all integers except 1

    S-A = {0}
    S-B = {1}

    ##A\cap B## = all integers except 0 and 1
    ##A\cap B## ≠ ∅.

    Also, you don't need any assumptions that look anything like that to prove that limits with respect to a metric are unique, but you probably know that. ##x_n\to x## means that every open ball around x contains all but a finite number of terms of the sequence. So if ##x_n\to x##, ##x_n\to y## and x≠y, just define r=(1/2)d(x,y) and consider the open balls B(x,r) and B(y,r). Since all but a finite number of terms are in B(x,r), and since B(y,r) is disjoint from B(x,r), B(y,r) contains at most a finite number of terms. This contradicts the assumption that ##x_n\to y##.

    Edit: Apparently I can't even read today. thought your post said =∅, not ≠∅.
     
    Last edited: Mar 27, 2012
  4. Mar 27, 2012 #3
    Re: "All but finitely many" theorem

    Hint:

    [tex]S\setminus (A\cap B)=????[/tex]
     
  5. Mar 27, 2012 #4
    Re: "All but finitely many" theorem

    Ah, I see, thanks!

    [itex]S\setminus (A\cap B) = (S\setminus A)\cup (S\setminus B).[/itex]

    If [itex]A\cap B = \varnothing[/itex], then [itex]S\setminus (A\cap B) = S\setminus \varnothing = S[/itex]. So [itex](S\setminus A)\cup (S\setminus B) = S[/itex]. But [itex]|S|=\aleph_0[/itex], whereas [itex]S\setminus A[/itex] and [itex]S\setminus B[/itex] are each finite, and the union of two finite sets is finite. Contradiction. Therefore [itex]A\cap B \neq \varnothing[/itex].

    To see that the union of two finite sets is finite, let [itex]P,Q[/itex] be finite. [itex]P\cup Q = P\cup (Q\setminus P)[/itex]. As a subset of a finite set, [itex]Q\setminus P[/itex] is finite. As a http://planetmath.org/encyclopedia/CardinalityOfDisjointUnionOfFiniteSets.html [Broken], [itex]|P\cup (Q\setminus P)| = |P| + |Q\setminus P|[/itex]. So [itex]|P\cup Q|=|P\cup (Q\setminus P)|=|P| + |Q\setminus P| \in \left \{ 0 \right \}\cup \mathbb{N}[/itex].
     
    Last edited by a moderator: May 5, 2017
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