Existence of a simple graph following some conditions Suppose, $a_1 < a_2 < \cdots < a_k$ are distinct positive integers. I am trying to prove that there exists a simple graph with $(a_k + 1)$ many vertices, whose set of distinct vertex degrees is $a_1, a_2, \cdots , a_k$.

I was trying to use induction on k. I solved using induction when $a_i = i , \forall i$.
But could not do the general case.
Any help will be appreciated. Thanks in advance.

A trick with complements helps. If a graph has $a_k+1$ vertices and the set of distinct degrees is $\{a_1, a_2, \dots, a_k\}$, then its complement is a graph with $a_k + 1$ vertices where the set of distinct degrees is $\{a_k-a_1, a_k - a_2, \dots, a_k - a_{k-1}, 0\}$.
The zero degrees are easy to tack on at the end, so now we have reduced to a smaller problem: find a graph with fewer than $a_k+1$ vertices where the set of distinct degrees is $\{a_k - a_1, a_k - a_2, \dots, a_k - a_{k-1}\}$. This can be done by induction.
