No Sidon sequence which is an asymptotic basis of order $2$ $\omega \subseteq \mathbb{R}^+$ is called a Sidon sequence,
if all the sums $a + a' \ (a, a' \in \omega, a \leq a')$ are distinct,
and it is an asymototic basis of order $2$,
if any positive integer $n$ sufficiently large can be expressed as a sum of $2$ elements of $\omega$.
According to the article I am reading, apparently it is not too difficult to show that there does not exist $\omega$ such that it is a Sidon sequence and also an asymototic basis of order $2$, but I am not quite seeing how to prove this at the moment. 
I was wondering if someone could possibly give me an explanation on how to show this? Thank you very much!
 A: Sidon set $A$ has at most $\sqrt{n}(1+o(1))$ elements not exceeding $n$ (*). So, $A+A$ contains at most $|A|(|A|+1)/2=n(1/2+o(1))$ elements not exceeding $n$, unlike an asymptotic basis of order 2. 
(*) may be proved as follows: fix $M$, denote your elements $x_1<x_2<\dots<x_m\leqslant n$ and consider all differences $x_j-x_i$ for $i<j\leqslant i+M$. They are all distinct, there areat least, say, $M(m-M)$ such differences and the sum of them does not exceed $\frac{M(M+1)}2 n$, thus $\frac{M(M+1)}2 n\geqslant (M(m-M))^2/2$, optimizing by $M$ or just by taking very large $M$ we get what we need: $m\leqslant \sqrt{n}(1+o(1))$.
A: An answer to your question can be found in this old paper by Erdős and Turán:
https://www.renyi.hu/~p_erdos/1941-01.pdf
In this paper, they also state their beautiful conjecture on additive bases:
if $B$ is a subset of the natural numbers and $f(n)$ represents the number of ways of writing $n$ as a sum of two elements in $B$, then $f(n) > 0$ for $n \geq n_0$ implies that $\limsup_{n \rightarrow \infty} f(n) = \infty$. This conjecture is still wide open.
