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.

Put $a_1$ vertices of degree $a_k$ and $a_k-a_{k-1}$ vertices of degree $a_1$.
The number of vertices remaining is $a_k + 1 - a_1 - (a_k-a_{k-1}) = a_{k-1}-a_1+1$.
We are reduced to the sequence :
$a_2 - a_1, a_3-a_1..., a_{k-1}-a_1$, so induction applies.
