Suppose $A_1 \subseteq A_2 \subseteq A_3 \subseteq\cdots,\,\,$ and suppose $\bigcup \limits_{n\in \mathbb{Z}^+} A_n = \mathbb{Z}^+$. Suppose $A_1 \subseteq A_2 \subseteq A_3 \subseteq\cdots,\,\,$
and suppose
$\bigcup \limits_{n\in \mathbb{Z}^+} A_n = \mathbb{Z}^+$.
Suppose also that for every infinite set B $\subseteq \mathbb{Z}^+$, there is some positive integer $n$ such that B$\cap A_n$ is infinite.
Prove that for some $n$, $A_n = \mathbb{Z}^+.$
I’m trying a contradiction argument.  So assume that for all n, $A_n \neq \mathbb{Z}^+$ . Then for every +ive integer i there is some $a_i \in \mathbb{Z}^+$ such that $a_i \notin A_i$. So from the givens, for any infinite B we can find j such that $A_j \cap B$ Is infinite but does not contain some $a_j$.
I don’t know where to go from here.  Maybe I need to think of a specific set to use for B, or perhaps a different approach entirely is required.  Any help would be greatly appreciated.
 A: Suppose that each $A_i$ is a strict subset of $\mathbb{Z}^+$. Moreover wlog suppose that the sequence of $A_i$ is strictly increasing (pass to such a subsequence otherwise). Then there is a sequence of $a_i$ such that $a_i\in A_i$ and $a_i\notin A_j$ for all $j<i$. This becomes an infinite sequence $B:=\{a_i\}_{i=2}^\infty$. But there is an $n$ such that $B\cap A_n$ is infinite. This is a contradiction since this must contain some $a_m$ with $m>n$.
A: Let $\mathbb{N}$ be equal to $\mathbb{Z}^+$.
We can prove this by using the contrapositive.
First, let's assume that $A_i = A_{i+1}$ if and only if $A_i = A_{i+1} = \mathbb{N}$.
If $A$ doesn't have this property, we can create a new sequence by dropping elements from our sequence that have appeared before.
Next, let's consider the contrapositive.
Suppose that there's no $n$ such that $A_n$ is equal to $\mathbb{N}$.
This means $A_\mathbb{N}$ satisfies the following:
$$ A_1 \subsetneq A_2 \subsetneq \cdots A_n \cdots $$
Since the inequalities are strict, this means, for every $n \ge 2$ there exists an $f(n)$ such that $f(n) \in (A_{n} \setminus A_{n-1})$.
Let's pick the set $B$ to be $\{ f(n) : n \ge 2 \}$.
The intersection of $B$ and any fixed $A_n$ is finite, but $B$ itself is infinite.
Thus we have shown
$$ \text{If there exists no $n$ such that $A_n$ is $\mathbb{N}$,} \\ \text{then there exists an infinite set $B$ such that $B \cap A_m$ is finite for all $m$} $$
And by contrapositive,
$$ \text{If there exists no infinite set $B$ such that $B \cap A_m$ is finite for all $m$,} \\ \text{then there exists an $n$ such that $A_n$ is $\mathbb{N}$} $$
