This is really stumping me.
I'm rewriting the problem so I'm not just asking for homework help, but I was studying for a test on logic and this problem came up:
Define an indexed family of sets $\{A_i\}_{i\in\mathbb{N}}$ where $A_i\subseteq\mathbb{N}$ for all $i\in\mathbb{N}$ that satisfies the following properties:
1. Every natural number is in at least one of the sets -> $(\forall n\in\mathbb{N})(\exists i \in \mathbb{N})(n\in A_i)$
2. None of the sets are equal to $\mathbb{N}$ -> $(\forall i\in\mathbb{N})(\exists n\in\mathbb{N})(n\not\in A_i)$
3. None of the sets have an upper bound -> $(\forall i \in \mathbb{N})(\forall m \in \mathbb{N})(\exists n\in\mathbb{N})(n > m \wedge n\in A_i)$
4. One of the sets is a strict subset of every other one -> $(\exists j\in\mathbb{N})(\forall i\in\mathbb{N})(i\neq j \rightarrow A_j\subsetneq A_i)$
I've been working on this for a good half hour. Does anybody know of a set that will fit those criteria?
Idea : take $A_1$ the set of all primes (which is unbounded). Takes $A_2$ the union of $A_1$ and the set of all primes times $2$. Takes $A_3$ to be the union of $A_1$ and the set of all primes times $3$. And so on for $A_k$.
By prime decomposition, every natural number $n$ is in $A_{n/p}$ for some prime $p$. Obviously by construction, $A_1$ is a proper subset of any $A_k$ for $k\ne 1$. Since $A_1$ is unbounded, so are the $A_k$'s. And no set are equal to $\mathbb N$. Consider $A_k$ and consider a prime $p > k$. Then $p^2 \cdot k \not \in A_k$.