Schirelman Theorem with even number set $E$ and its density modification

Question

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The above one is the excerpt from a book involving sieve theory introduction. I have a question about this one. Let $E=\left\{2n : n\in\mathbb N\right\}$ be a set of all even number. The book states that the Schirelman density of set $A$, $$d(A)=\inf_{n\ge 1}\dfrac{A(n)}{n}$$ where $A(n)=\#\left\{a\le n : a\in A\right\}$. We see that $d(E)=0$ because $E(1)=0$, yet actually $d(E)>1/3$ for sufficiently large $n$ ($2$ in this case is enough). The modification above, in my opinion that I'm not sure it's true is to modify the density to be $\displaystyle d(A)=\inf_{n\ge n_0}\dfrac{A(n)}{n}$ and Corollary 6.3.3 above is then states that if (modified) $d(A)>0\Longrightarrow$ $\exists m, A^{(m)}=\mathbb{N}_{\ge n_0}$, where $\mathbb{N}_{\ge n_0}$ is the set of all sufficiently large natural number.

However, since the modified density $d(E)>0$ for $n\ge 2$ and we know it's impossible to have such $m$ in which $E^{(m)}=\mathbb{N}_{n_0}$, since the sum of even is always even and not cover the case of odds, so I don't know what these sentences are all about.

** $A^{(n)}=\left\{a_1+a_2+\dots a_n : a_i\in A\right\}\cup A$

Answer 1

The passage you quote does leave some room for interpretation. Before discussing it, let's set the following notation. Given a natural number $n_0$, set $d_{n_0}(A)=\inf_{n\geq n_0}A(n)/n$.

You have interpreted the passage as follows.

Statement 1. Suppose there is some natural number $n_0$ such that $d_{n_0}(A)>0$. Then there is some $m$ such that $A^{(m)}$ contains all sufficiently large natural numbers.

As you point out, the set of even numbers does refute Statement 1. However, my guess is that the author actually means the following:

Statement 2. Suppose there are natural numbers $n_0$ and $k$ such that $d_{n_0}(A^{(k)})>1/2$. Then $A^{(2k)}$ contains all sufficiently large natural numbers.

So first note that the set $E$ of evens is no longer a counterexample, because $d_{n_0}(E^{(k)})\leq 1/2$ for all $n_0$. In fact, Statement 2 is true.

Proof of Statement 2. We prove the general claim that if $d_{n_0}(A)>1/2$ then $A+A$ contains all natural numbers greater than $n_0$. So assume $d_{n_0}(A)>1/2$. Then $A(n)>n/2$ for all $n\geq n_0$. Now fix $n\geq n_0$. Set $X=\{a\in A:1\leq a\leq n\}$ and $Y=\{n+1-a:a\in X\}$. Then $X$ and $Y$ are subsets of $[1,n]$ of size $A(n)>n/2$. So $X$ and $Y$ intersect, i.e., $n+1-a=b$ for some $a,b\in A$, i.e., $n+1\in A+A$.

Now the natural question is how close we can get to something like Statement 1. One option is the following.

Theorem. Let $r=gcd(A)$ and assume $d_{n_0}(A)>0$ for some $n_0$. Then there is some $k$ such that $A^{(k)}$ contains all sufficiently large multiples of $r$.

Note that since every element of $A$ is a multiple of $r$, this is optimal.

The previous theorem is a consequence of Lemma 1 of Cofinite subsets of asymptotic bases for the positive integers, J. Number Theory 1985 by John C. M. Nash and Melvyn B. Nathanson. The proof sketch is as follows. Without loss of generality, you can replace $A$ by $\{a/r:a\in A\}$ and thus assume $r=1$. Let $B=A\cup\{1\}$. Then $d_{n_0}(A)>0$ implies $d(B)>0$ (this part doesn't use $r=1$). So by Corollary 6.3.3 there is some $m$ such that $B^{(m)}=\mathbb{N}$. Then there are some tricks using the assumption $gcd(A)=1$ to get from $B^{(m)}=\mathbb{N}$ to $A^{(k)}$ containing sufficiently large natural numbers for some $k$.