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.


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.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.