Given a topological sorting $>$ over a set of elements $N$, how many possible DAGs with vertex set $N$ that are consistent with $>$ ? ( note that a DAG may be consistent with other orders in addition to $>$ )
If it is not known, can we bound it by some number?
My attempt: Assume $x_1>x_2>\dots>x_n$ to be the given order. For the first element $x_1$, we can connect it to any (subset) of the rest. Thus, for the element at position $i\in [1,2,\dots,n]$ we can connect it to $\sum\limits_{k=0}^{n-i} \binom{n-i}{k}$ other elements and finally we have $\sum\limits_{i=1}^{n}\sum\limits_{k=0}^{n-i} \binom{n-i}{k}$ possible DAGs over $>$.
Does this give the correct number for it?
There are $\binom{n}2$ pairs of vertices. Any pair can be connected, and if it is connected, the direction is unique. Thus, there are $2^{\binom{n}2}$ possible DAGs consistent with a given linear order on $[n]$, one for each subset of the pairs of vertices.
Your calculation simplifies to
$$\sum_{i=1}^n\sum_{k=0}^{n-i}\binom{n-i}k=\sum_{i=1}^n2^{n-i}=\sum_{i=0}^{n-1}2^i=2^n-1\;,$$
which is clearly too small. The problem is that the outer summation should be a product: each combination of connections for one vertex can be combined with any combination of connections for a later vertex. Then you get the correct value:
$$\prod_{i=1}^n\sum_{k=0}^{n-i}\binom{n-i}k=\prod_{i=1}^n2^{n-i}=\prod_{i=0}^{n-1}2^i=2^{\sum_{i=0}^{n-1}i}=2^{\binom{n}2}\;.$$