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In a paper I am reading (lemma 6) the author uses without proof that there exists a family of subsets $\{A_i\subset \mathbb{N}\}_{i\in \mathbb{R}}$ s.t. $A_i\Delta A_j$ is infinite and co-infinite for $i\not=j$. ($\Delta$ is the symmetric difference)

If the index set were countable, this is obvious. We can take for example $A_i=\{n p_i\mid n\in \mathbb{N}\}$ where $\{p_i\}$ is some enumeration of the primes. However, for an uncountable index set I'm not sure we can explicitly write down such a family of subsets, and guess that we probably have to use the axiom of choice.

Even using the axiom of choice I'm not sure how to construct such a family though, and so I'm any suggestions as to how to construct such a family.

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Let $E=\{2n:n\in\Bbb N\}$, and let $\mathscr{B}=\{B_x:x\in\Bbb R\}$ be an almost disjoint family of infinite subsets of $E$; you’ll find constructions in the answer to this question and the one to which it’s linked as a duplicate. For $x\in\Bbb R$ let $A_x=B_x\cup(\Bbb N\setminus E)$; then $\mathscr{A}=\{A_x:x\in\Bbb R\}$ has the desired properties.

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  • $\begingroup$ What if $B_x,B_y\subset (\mathbb{N}\setminus E)$? Then the symmetric difference $A_x\Delta A_y=\emptyset$. $\endgroup$ – user2520938 Jun 19 '16 at 10:43
  • $\begingroup$ @user2520938: That’s impossible: the sets $B_x$ are by definition subsets of $E$. $\endgroup$ – Brian M. Scott Jun 19 '16 at 10:48
  • $\begingroup$ Oh sorry, should have read more closely. Really nice construction, thank you $\endgroup$ – user2520938 Jun 19 '16 at 10:50
  • $\begingroup$ @user2520938: You’re welcome. $\endgroup$ – Brian M. Scott Jun 19 '16 at 10:50

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