Using Zorn's Lemma-Topology

Question

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.

Answer 1

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 $\subseteq$.

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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.