I was trying to solve the following problem from Tao-Vu.
Let $A$ be any additive set. Show that a Sidon set contained in $A$ can have cardinality at most $\sqrt{2\sigma[A]|A|}.$
Suppose that $S$ is a Sidon set in $A$. Then the following set-theoretical equality is valid and the union is disjoint $$S\times S=\bigsqcup_{t\in S+S} \{(x,y)\in S\times S: x+y=t\}.$$
Since $S$ is a Sidon set, then if $\{a,b\}\neq \{c,d\} \Rightarrow a+b\neq c+d.$ Hence for any $t\in S+S$ the set $\{(x,y)\in S\times S: x+y=t\}$ has exactly two elements. Therefore $|S|^2=\sum \limits_{t\in S+S}\#\{(x,y)\in S\times S: x+y=t\}=\sum \limits_{t\in S+S}2=2|S+S|.$ Since $S+S\subset A+A,$ then $|S|^2\leq 2|A+A|$ and if we take square root we obtain desired inequality because $\sigma[A]:=\dfrac{|A+A|}{|A|}.$
Firstly I thought that this equality $|S|^2=2|S+S|$ is valid, if $S$ is a Sidon set but later I realized that it is false. If $S$ is a Sidon set then $\sigma[S]:=\dfrac{|S+S|}{|S|}$ is maximal and equal to $\dfrac{|S|+1}{2}$. But our equality implies that $\sigma[S]=\dfrac{|S|}{2}$ and this is contradiction. So something is wrong.
What is my mistake? I could not find my mistakeю
You write:
But this is not correct; if $t=a+a$ for some $a\in S$ then the corresponding set has only one element. Your idea is however correct; the correct computation is that \begin{align} |S|^2&=\sum_{t\in S+S}\#\{(x,y)\in S\times S:x+y=t\} \\ &=|S|+2\cdot(|S+S|-|S|) \\ &=2|S+S|-|S|. \end{align} We thus still get the inequality $|S^2|\leqslant 2|S+S|\leqslant 2|A+A|$, so your proof still carries through.