If $A,B \subset \mathbb N$ with $\delta(A)=\delta(B)=0$, does $\delta(A\cup B)$ exist?

Question

I know that every subset of $A\subset\mathbb N$ may not have asymptotic/natural density $\delta(A)$. The limit exists if the limsup and liminf are equal.

In this context, I'm finding some results on the existence of natural density of any subset of $\mathbb N$.

Particularly I need to know the followings : If $A,B \subset \mathbb N$ with $\delta(A)=\delta(B)=0$, does $\delta(A\cup B)$ exist?

Answer 1

Yes, $\delta(A \cup B)$ exists and equals zero.

Let $J_n = \{1,2,\dots,n\}$, and simply note that $|(A \cup B) \cap J_n| \le |A \cap J_n| + |B \cap J_n|$ (it would be equality if $A,B$ were disjoint). Hence $$\frac{|(A \cup B) \cap J_n|}{n} \le \frac{|A \cap J_n|}{n} + \frac{|B \cap J_n|}{n}.$$ As $n \to \infty$, both terms on the right side approach 0 by assumption, hence the left side must also, and this says precisely that $\delta(A \cup B) = 0$.

Answer 2

Yes. But note that it doesn't have to if you just assume that $\delta(A)$ and $\delta(B)$ exists. Indeed, let $A$ be the set of even integers and define $J: \mathbb{N} \rightarrow \mathbb{N}$ at an $n$ by letting $k$ be the unique integer such that $2^k\leq n<2^{k+1}$ and let $J(n)$ be $2n$ if $k$ is even and $2n-1$ if $k$ is odd. Putting $B:=J(\mathbb{N})$ we see that $|(A\cup B)\cap\{1,\cdots ,n\}|/n$ will in fact oscillate between $2/3$ and $5/6$.

Answer 3

Yes, and in fact you don't have to assume that both densities are equal to $0$, just one of them.

Let $[n]=\{1,\dots,n\}$.

Since $$\frac{|(A\cup B)\cap[n]|}n=\frac{|A\cap[n]|}n+\frac{|B\cap[n]|}n-\frac{|(A\cap B)\cap[n]|}n$$ by the in-and-out principle, it follows that $\delta(A\cup B)$ exists whenever $\delta(A)$ and $\delta(B)$ and $\delta(A\cap B)$ all exist, and in that case $\delta(A\cup B)=\delta(A)+\delta(B)-\delta(A\cap B)$.

In particular, if $\delta(A)$ exists and $\delta(B)=0$, then $\delta(A\cup B)$ exists, and $\delta(A\cup B)=\delta(A)$.