Find $A$ and $B$ such that $A⊈B$ and $B⊈A$?

Question

I need to prove that the subset relation “$⊆$” on all subsets of $\mathbb Z$ is not a total order and I'm going to do this by finding $A$ and $B$ such that $A⊈B$ and $B⊈A$?

Is there a simple solution for this?

Answer 1

Yes. For instance $$ A = \{1\}, B = \{2\}\\ A = \text{All the negative integers},B = \text{All the positive integers}\\ A = \{2n \mid n \in \Bbb Z\}, B = \{2n+1\mid n \in \Bbb Z\}\\ A = \Bbb Z \setminus \{1\}, B = \Bbb Z \setminus \{2\}\\ A = \{5n \mid n \in \Bbb Z\}, B = \{3n \mid n \in \Bbb Z\} $$

Answer 2

Any two subsets with a non-empty symmetric difference ($A \setminus B) \cup (B \setminus A$) satisfy your requirements.

Answer 3

Yes. For instance,

$A=n\mathbb{Z}$ and $B=m\mathbb{Z}$, with $gcd(m,n)=1.$

Answer 4

Example of subsets can be chosen from the smaller sets $\{1,2,\ldots 30\}$. Let $A$ be the set of dates in a calendar month when Alice is going to have pizza for dinner, and let $B$ be the set of dates Bob is going to have pizza.

Now $A\subset B$ means, on all the dates Bob had pizza Alice had it too.

Now you can draw up a pizza eating schedule for both of them meeting the condition in your question.