My uni is conducting a short lecture series on Additive Combinatorics (it's mostly concerned about sumsets and Number Theory) whence I came across Dyson's transform and Cauchy-Devenport theorem.
$\textbf{Notation}:$ Let $A$ be a sequence of natural numbers and $x\in\mathbb R$. Define: $A(x)= \left|A\cap[1, x]\right|\tag*{}$
Suppose, $A=\{a_1<a_2<\cdots\}\tag*{}$ and $B=\{\underbrace{b_1}_{=0}<b_2<\cdots\}\tag*{}$ are two sequences of $\mathbb N\cup\{0\}$. As usual, $A+B=\{a+b \ : \ a\in A, \ b\in B\}\tag*{}$ and $A-B=\{a-b \ : \ a\in A, \ b\in B\}\tag*{}$
Let $e\in A$ and $m\in\mathbb R$.
Consider the transformed sequences over $e$: $$A'=A\cup (B+\{e\})\\ B'=B\cap(A-\{e\})$$
These are the properties I am attempting to prove:
- $A'+B'\subset A+B$
- $|A'|+|B'|=|A|+|B|$
- $A'\setminus A =\{e\}+B\setminus B'$
- $0\in B'$
- $A'(m)+B'(m-e)=A(m)+B(m-e)$
My attempt:
$(4)$ is fairly straight forward. $0\in B$ (by defn.) and $e\in A\implies 0\in A-\{e\}$. As defined, $B'=B\cap(A-\{e\})$, thus, $0\in B'$. $\blacksquare$
$(2)$. $|B'|=|B\cap (A-\{e\})|\tag*{}$
$|A'|=|A|+|B+\{e\}|-|A\cap (B+\{e\})|\tag*{}$
Claim: $|B|=|B+\{e\}|$ and $|B\cap (A-\{e\})|=|A\cap (B+\{e\})|$
Proof: Define $f(b)=b+e$. $f:B\to B+\{e\}$, $f:B\cap (A-\{e\})\to A\cap (B+\{e\})$ are one-one.
$\therefore |A'|+|B'| = |A|+|B|\ \blacksquare$
$(3).$ $A'\setminus A=(A\cup(B+\{e\}))\setminus A \\ =(B+\{e\})\setminus A\tag*{}$ and $B\setminus B' = B\setminus (B\cap (A-\{e\})) \\= B\setminus(A-\{e\})\tag*{}$ I think now it's obvious to see that $B\setminus B' +\{e\}= A'\setminus A$ $\blacksquare$
Not sure if these are rigorous enough, it makes sense to me. However, I am not able to prove the remaining two i.e., $(1)$ and $(5)$. Can anyone help me out or suggest resources from where I can learn this topic i.e., Dyson's transform?