# Zorn's lemma - I think I made a counterproof

This is from a textbook on topology:

A subset W of the set Z of integers is said to be closed under addition if given any elements w and w′ of W, w+w′∈W.

Prove that there is a maximal subset of Z which is closed under addition and does not contain 9

My problem is as follows: The set of all multiples of 4 does not contain 9 and is closed under addition, the same holds for all multiples of 5. A maximal set must contain both sets, but if that set is closed under addition, we get that 9 is a member of the set.

EDIT: Thanks for the clarification, on the definition I was really lost. Can anyone give me a sketch of a proof?

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The meaning of a maximal set $M$ among a collection $\mathcal A$ of subsets is that $M\in\mathcal A$ and $\forall A\in\mathcal A,A\supseteq M\implies A=M$, therefore you cannot conclude that $M$ contains $4\mathbb Z,5\mathbb Z$, etc. –  Frank Science Feb 16 '14 at 6:51
I'm not well versed in set theory, but what does this have to do with Zorn's lemma? –  Jack M Feb 16 '14 at 10:41
The point is that the proof of this statement uses Zorn's lemma. –  hunter Feb 16 '14 at 11:41
Actually I take it back - you can show by hand that all even numbers form a maximal set. –  hunter Feb 16 '14 at 11:43
Isn't Z itself the only maximal set according to this definition ? It's definitely closed and definitely maximal (but does contain 9). –  Tom Collinge Feb 16 '14 at 14:09

Maximal set means that you just cannot add more elements while preserving this property, so indeed a maximal set might include the multiples of $4$ or the multiples of $5$, but certainly not both as you show.
(Recall that if $(P,\leq)$ is a partially ordered set, a maximal element $p\in P$ is such that whenever $p\leq q$ we have that $p=q$; whereas a maximum is an element $p$ such that for every $q$ we have $q\leq p$.)
In other words his argument with $4\mathbb{Z}\in\mathcal{A}$ and $5\mathbb{Z}\in\mathcal{A}$ does prove that the collection $\mathcal{A}$ of additively closed sets excluding $9$, cannot have a maximum $G\in\mathcal{A}$. But that does not rule out the possibility of maximal elements $M\in\mathcal{A}$. –  Jeppe Stig Nielsen Feb 16 '14 at 10:54
@Tom: We're only interested in sets that $9$ is not their element. Clearly $\Bbb Z$ is out of the question. If you talk about "subsets of $\Bbb Z$" and you talk about "Subsets which have a particular property that $\Bbb Z$ itself does not posses", then yes you talk about proper subsets. –  Asaf Karagila Feb 17 '14 at 8:20
@Tom: And that would be very poorly worded, since there is a maximum amongst the sets not including $9$, and it is certainly not closed under addition. –  Asaf Karagila Feb 17 '14 at 14:27