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Question:

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

Prove that there is a maximal subset of $Z$ which is closed under addition and does not contain $9$. Do this using Zorn's lemma.

Attempt:

Consider $S=\lbrace(-n,n)\mid n\in Z-\{9\} \rbrace $. Then for all $n$, $(-1,1)\subset (-2,2) \subset (-3,3)\ldots$ Hence, $S$ is totally ordered. For any $s\in S$, there exists $s'\in S$ where $s'=\lbrace -n-1,n+1 \rbrace$ is an upper bound.

Thus, by Zorn's lemma, $S$ contains a maximal element.

Comments:

Is this correct? All help is appreciated.

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    $\begingroup$ I just wonder why you were asked to use Zorn's lemma for this question. The set $S = 2\mathbb{Z}$ is a semigroup not containing $9$ and it is maximal for this property. Indeed, let $T$ be semigroup containing $S$ and let $x \in T \setminus S$. Then $9 - x$ is in $S$ and hence in $T$. Thus $9 = (9 - x) + x$ is in $T$. No Zorn's lemma... $\endgroup$ – J.-E. Pin Sep 9 '13 at 7:37
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You should be applying Zorn's Lemma to the following partially ordered set:

  • the underlying set is the family $\mathcal{P}$ of all subsets of $\mathbb{Z}$ which are closed under addition, and do not contain $9$;
  • the ordering is the subset relation $\subseteq$.

In particular, the outline of your proof should be as follows:

Suppose that $\mathcal{C}$ is a chain in (totally ordered subset of) the partial order $\mathcal{P}$ described above. (Note that this means that each element of $\mathcal{C}$ is a subset of $\mathbb{Z}$ closed under addition which does not contain $9$, and that for any two $B_1,B_2$ in $\mathcal{C}$ either $B_1 \subseteq B_2$ or $B_2 \subseteq B_1$.) [Do some work to show that there is an element $A$ of $\mathcal{P}$ such that $B \subseteq A$ for each $B \in \mathcal{C}$; there is an obvious candidate which will work.] Conclude that the partial order $\mathcal{P}$ satisfies the hypothesis of Zorn's Lemma, and therefore the partial order has a maximal element.

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  • $\begingroup$ Thanks for the hint. Will reform my solution. $\endgroup$ – Julius Jackson Sep 9 '13 at 5:16

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