A partial order with specified properties I will say that two elements $a$ and $b$ of a poset are intersecting ($a\not\asymp b$) when there exists non-least element $c$ such that $c\le a$ and $c\le b$.
Let $\ge_0$ be a partial order on a set $\mho$ (we may additionally require it to be a distributive lattice or whatever if this helps). Let $N$ be an index set.
Is there a partial order $\ge_1$ on $\mho^N$ such that
$$X\not\asymp_1 Y \Leftrightarrow \forall i\in N:X_i\not\asymp_0 Y_i.$$
If such a partial order doesn't exist, which kind of relation it can be? (maybe a preorder or whatever?)
 A: For $a,b$ in a poset $P$ with partial order $\le$, let $$\text{prec}^P_{\le}(a,b):=\{c\in P:c\le a,c\le b\}.$$ Then $a\not\asymp b$ if and only if there is some non-least element of $P$ in $\text{prec}^P_{\le}(a,b)$. If $P$ has no least element, then $a\not\asymp b$ if and only if $\text{prec}^P_{\le}(a,b)\ne\emptyset$.
Now, let's suppose that $\mho$ has no $\le_0$-least element (I'll think on the case where it does). Let's define $\le_1$ on $\mho^N$ by $X\le_1 Y$ iff for all $i\in N$, $X_i\le_0 Y_i$. Then $\le_1$ is readily seen to be reflexive, transitive, and antisymmetric (since $\le_0$ is). Note moreover that $\mho^N$ has no $\le_1$-least element, since $\mho$ has no $\le_0$-least element.
Now, for any $X,Y\in\mho^N$, we have
$$\begin{align}X\not\asymp_1 Y &\Longleftrightarrow \exists Z\in\text{prec}^{\mho^N}_{\le_1}(X,Y)\\ &\Longleftrightarrow \exists Z\in\mho^N:(Z\le_1 X)\wedge(Z\le_1 Y)\\ &\Longleftrightarrow \exists Z\in\mho^N:\: \forall i\in N(Z_i\le_0 X_i)\wedge(Z_i\le_0 Y_i)\\ &\Longleftrightarrow \exists Z\in\mho^N: \forall i\in N (Z_i\in\text{prec}_{\le_0}^\mho(X_i,Y_i))\\ &\overset{*}\Longrightarrow \forall i\in N\:\exists Z_i\in\text{prec}_{\le_0}^\mho(X_i,Y_i)\\ &\Longleftrightarrow \forall i\in N\: (X_i\not\asymp_0 Y_i).\end{align}$$ The converse of the starred implication holds for all index sets $N$ if (and I suspect only if) the Axiom of Choice for subsets of $\mho$ does. (Of course, if you are fine using the Axiom of Choice, then this is no problem for you.)
Edit
We can extend the above to the cases where $N=\emptyset$, or $\mho$ is a singleton, for in either case we have $\mho^N$ a singleton, and the only non-empty relation on $\mho^N$ will necessarily satisfy the desired properties. (Why?) Also, if $\mho=\emptyset$ and $N\ne\emptyset$, then $\mho^N=\emptyset$, and the only relation on $\mho^N$ vacuously satisfies the desired properties. Finally, if $N$ is a singleton, then $\mho^N$ is effectively just $\mho$, and the desired relation is readily defined.
Note that for any poset $\langle P,\le\rangle$ and $a\in P,$ we have $a\not\asymp a$ if and only if $a$ is not the $\le$-least element of $P$.
Suppose that $\mho,N$ have more than one element and that $\mho$ has a $\le_0$-least element. Then there is no partial order $\le_1$ on $\mho^N$ with the desired property. Indeed, take $j,k\in N$ distinct, let $x_0$ be the $\le_0$-least element of $\mho$, and let $x$ be any other element of $\mho$. Define $X,Y\in\mho^N$ by $$X_i=\begin{cases}x_0 & i=j\\x & \text{otherwise}\end{cases}$$ and $$Y_i=\begin{cases}x_0 & i=k\\x & \text{otherwise.}\end{cases}$$
Since $j\ne k$, then $X\ne Y$, so for any partial order $\le_1$ on $\mho^N$, we know that at least one of $X,Y$ is not the $\le_1$-least element of $\mho^N$ (there may be no $\le_1$-least element at all, but that doesn't matter). Without loss of generality, suppose that $X\not\le_1 Z$ for some $Z\in\mho^N$. Then $X\not\asymp_1 X$, but it is not the case that $X_j\not\asymp_0X_j,$ since $X_j=x_0$ is the $\le_0$-least element of $\mho$.

I will think more about how we may relax the requirements on $\le_1$ to a preorder or something else, but before I can answer that definitively, I'll need you to answer the following:

In the context of a set $X$ with a relation $\le$ that isn't a partial order on $X$, what do you want to mean by "a non-least element of $X$"? Do you simply mean some $x\in X$ such that $\exists y\in X:x\not\le y$? If $\le$ is in particular not antisymmetric, and there exist distinct $x_1,x_2\in X$ such that $\forall y\in X((x_1\le y)\vee(x_2\le y))$, would you say that $x_1,x_2$ are both $\le$-least elements of $X$, or that $X$ has no $\le$-least elements at all?

If I don't know that, I can't very well figure out what $\not\asymp$ means in general.
