$\DeclareMathOperator\inc{inc}$Let $|X|=n$ and $\inc(X,\leq)=\{\{x,y\} : \neg (x\leq y)\wedge \neg (y\leq x)\}$, where $(X,\leq)$ is poset (possibly unconnected). Define the function:
$$\pi(n,m):=|\{(X,
\leq):\inc(X,\leq)=m\}/\cong|$$
, where $\cong$ is the relation of isomorphism of partial orders. It's obvious that if $m>\binom{n}{2}$, then $\pi(n,m)=0$. Is it true that for any $m\in [0,\binom{n}{2}]$ $\pi(n,m)>0$? I know that if $\sum\_{i=1}^k a\_i=n$ then
$$\pi\left(n,\sum\_{i=1}^{k-1}\left(a\_i\sum\_{j=i+1}^{k}a\_j\right)\right)>0$$
and if $1<m\leq n$ then $\pi(n,\binom{m}{2})\geq n-m+1$. I'm also interested in upper and lower bounds for $\pi(n,m)$. For example, if $n-3m+1=k>0$ then $\pi(n,m)\geq k$ . Maybe this MO question is related: [Poset of antichains of given cardinality](https://mathoverflow.net/questions/371744/poset-of-antichains-of-given-cardinality) .



The quantity of poset with a given number of pairs of incomparable elements
Yes. This is true. We shall prove this result by induction on $n$. Suppose that $n>0$. and $0\leq m\leq\binom{n}{2}$. If $m\leq\binom{n-1}{2}$, then there is some poset $X$ with $|X|=n-1$ and where $\text{inc}(X)=m$. Now attach a new element $1$ where $x\leq 1$ for $x\in X$ to obtain a poset $X\cup\{1\}$. Then $|X\cup\{1\}|=n$ and $\text{inc}(X)=m.$ Therefore, $\pi(n,m)>0$.


Let $r\leq\binom{n-1}{2}$. Suppose now that $|X|=n-1$ and $\text{inc}(X)=r$. Then let $c$ be a new element that is incomparable with all the elements in $X$. Then $|X\cup\{c\}|=n$ and $\text{inc}(X\cup\{c\})=r+n-1$. In particular, if we have $\binom{n}{2}\leq m\geq n-1$, then we can set $r=m-(n-1)$, so that $|X\cup\{c\}|=n$ and $\text{inc}(X\cup\{c\})=m$. We have covered the cases when $\binom{n}{2}\geq m\geq n-1$ and when $m\leq\binom{n-1}{2}$, and one can easily verify that this covers all possible cases $m$ with $0\leq m\leq\binom{n}{2}$.


**Alternate proof**


The result also follows from the one-point extension property of partial orders.


Theorem: Suppose that $(X,P)$ is a partially ordered set with $X$ finite. Let $(X,Q)$ be a partially ordered set with $P\subseteq Q,P\neq Q$. Then there exists a pair $(x',y')\in Q\setminus P$ where $P\cup\{(x',y')\}$ is a partial ordering.


Construction of $(x',y')$: Let $(x,y)\in Q\setminus P$. Suppose now that $(x',y')$ is a pair with $x'\leq\_P x,y\leq\_P y',x'\not\leq\_P y'$ and where
$$x''\leq\_P x',y'\leq\_P y''\Rightarrow(x''\leq\_P y''\,\text{or}\,x''=x',y''=y').$$


I gave a proof that $P\cup\{(x',y')\}$ is actually a partial order in [my other answer to another question here](https://mathoverflow.net/a/418100/22277).


Since every partial order can be extended to a linear order, for each $n$ and set $X$ with $|X|=n$, we have a sequence of partial orders $P\_0\subseteq\dots\subseteq P\_m$ where $m=\binom{n}{2}$ where $P\_0=\{(x,x)|x\in X\}$, and where $P\_m$ is a linear order, and where $|P\_{i+1}|=|P\_i|+1$ so that $|P\_i|=n+i$. In this case, $\text{inc}(X,P\_i)=m-i$.


