I've been thinking about the Schnirelmann Density and I think that I may still be confused about SumSet and Density.
It seems to me that if $d(A) \ge \frac{1}{2}$ and $d(B) > 0$, then $d(A+{B}) = 1$.
Here's my thinking:
(1) if $1 \notin A$, then $d(A)=0$ so if $d(A) \ge \frac{1}{2}$, then $1 \in d(A)$
(2) if $x \in A$, then $x+2 \in A$, otherwise $d(A) < \frac{1}{2}$
(3) So if $C = \{ \text{odd integers} \}$, that is, $C = \left\{1, 3, 5, 7, \dots\right\}$, then $C \subseteq A$
(4) But using the sumset, $A+\{1\}$ includes all even numbers.
(5) So, it follows that if $d(B) > 0$, then $1 \in B$, and it follows that $d(A+B)=1$.
Am I correct? Is there a mistake in my reasoning?
Here's my summary of my understanding of Schnirelmann density:
$A,B$ are infinite sequences of integers starting with $0$ with in sequential order such as $0, a_1, a_2, \cdots$ where $0 < a_1 < a_2 < \cdots$
Schnirelmann density is defined as: $$d(A) = \inf\limits_{n}\frac{A(n)}{n}$$
where: $$A(n) = \sum\limits_{0<a_i\le{n}}{1}$$
- So, it is clear that: $$0 \le \frac{A(n)}{n} \le 1$$
My mistake is step#2. For example if we have $A =\{1, 2, 5, 7, 9, \dots \}$, we still have $d(A) = \frac{1}{2}$.