Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
It appears to me that you are unclear about how the outline of a proof will look. Please see the addition to my answer below. –  Arthur Fischer Sep 9 '13 at 7:26
1  
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... –  J.-E. Pin Sep 9 '13 at 7:37
add comment

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
Thanks for the hint. Will reform my solution. –  Julius Jackson Sep 9 '13 at 5:16
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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