Let $P_{\aleph_{0}}(\mathbb{N}) = \{x\in P(\mathbb{N}):|x|=\aleph_{0}\} $
we'll define the following relation on $P_{\aleph_{0}}(\mathbb{N})$ for $A, B \in P_{\aleph_{0}}(\mathbb{N})$:
$A\preccurlyeq B$ if for every $b_1, b_2 \in B \longrightarrow\left(b_1 < b_2 \longrightarrow\vert [b_1, b_2] \cap A\vert\geq2\right)$
Prove: $ \left\langle \preccurlyeq,P_{\aleph_{0}}(\mathbb{N})\right\rangle $ is a Partially ordered set.
my attempt:
reflexivity:
let $x_1, x_2 \in X$ suppose $x_1 < x_2 \longrightarrow \vert[x_1,x_2]\vert \geq 2, \ [x_1, x_2] \cap X = [x_1, x_2]$ - and wer'e done.
I'm confused about antisymmetry and transitivity.
intutiviely, it's clear why $A\preccurlyeq B \land\ B\preccurlyeq A\longrightarrow A=B$ but i'm not sure how to finish the proof. what is the right way to argue if for every $ a_1, a_2 \in A$ such that $a_1 < a_2 \longrightarrow\vert[a_{1},a_{2}]\cap B\vert\geq2$ and for every $ \ b_1, b_2 \in B $ such that $b_1 < b_2 \longrightarrow\vert[b_{1},b_{2}]\cap A\vert\geq2 $ then $A = B$?
same problem for transitivity.
I don't know what technich to use in order to prove the claim.
For antisymmetry, assume $A\leq B$ and $B\leq A$, and toward a contradiction assume that e.g. $A\nsubseteq B$ and let $a\in A\smallsetminus B$. Let $a'$ be the minimum of $\{x\in A\mid a<x\}$. By $B\leq A$ you can find $b,b'$ such that $a<b<b'\leq a'$ ($a<b$ holds since $a\notin B$). But now, by the choice of $a'$, $[b,b']\cap A$ contains at most one element (it is $a'=b'$ if they are equal) contradicting $A\leq B$.
Transitivity is straight-forward. If $A\leq B\leq C$, choose $c<c'$ in $C$. By $B\leq C$ you can find $b,b'\in B$ with $c\leq b<b'\leq c'$, and then by $A\leq B$ you can find $a,a'\in A$ with $b\leq a<a'\leq b'$. This is it, $[c,c']\cap A$ contains at least $a$ and $a'$.