Let $d \in P$. Show that the following two conditions on a finite poset $P$ are equivalent:
i. $P$ is the intersection of d linear orderings of $[n]$, where $|P| = n$,
ii. $P$ is isomorphic to a subposet of $\mathbb{N}^d$
Moreover, show that when $d=2$ the two conditions are also equivalent to:
iii. There exists a poset Q on [n] such that $s<t$ or $s>t$ in Q if and only if s
and t are incomparable in P.
So I know that I need to show $i \equiv ii$ and then either $i \equiv iii$ or $ii \equiv iii$, since the first two are already equivalent.
Here are my thoughts:
$\mathbb{N}^d$ is $(x_1,x_2,...,x_d)\leq (y_1,y_2,...,y_d)$ iff $x_i\leq y_i, \forall i\in [d]$
The intersection of $d$ linear orderings of $[n]$ is a new poset that $i\leq j$ iff $i\leq j$ in all $d$ linear orderings.
But I'm not sure how to show that $i. \equiv ii.$, so I can't even get started on showing the equivalence to $iii.$
Any insight, next steps, or guides on how to show these equivalences would be appreciated.
HINT: To show that (i) implies (ii), let $P$ be a partial order on $[n]$ such that $P=\bigcap_{k=1}^d\le_k$, where each $\le_k$ is a linear order on $[n]$. For each $k\in[d]$ there is a bijection $\varphi_k:[n]\to[n]$ such that for all $i,j\in[n]$, $\varphi_k(i)\le\varphi_k(j)$ iff $i\le_k j$. Let
$$\varphi:[n]\to[n]^d:i\mapsto\langle\varphi_1(i),\varphi_2(i),\ldots,\varphi_d(k)\rangle\,,$$
and show that $\varphi$ is an isomorphism. For the reverse implication you have to start with an isomorphism $\varphi$ and use it to define the linear orders $\le_k$ for $k\in[d]$; this is really just reversing the procedure that I just sketched.