Suppose we index subsets of the natural numbers in the following way.
$$\begin{matrix} X_2 = 4 \mathbb{N} &Y_2 = 8\mathbb{N} \\ X_3 = 16\mathbb{N} & Y_3 = 32\mathbb{N} \\ X_4 = 64\mathbb{N} & Y_4 = 128\mathbb{N} \\ \vdots & \vdots \end{matrix}$$
so that generally we have $X_n = 4^{n-1}$ and $Y_n = 2 \cdot 4^{n-1}$. I am wondering how to describe the set
$$\displaystyle\bigcup_{n=2}^\infty X_n \setminus Y_n$$
Clearly, we have
$$\displaystyle\bigcup_{n=2}^\infty X_n \setminus Y_n = (4\mathbb{N} \setminus 8 \mathbb{N}) \cup (16 \mathbb{N} \setminus 32 \mathbb{N}) \cup \dots$$
Since $8 \mathbb{N} \supset 16\mathbb{N} \supset 32\mathbb{N} \supset ...$, as soon as we remove $8\mathbb{N}$, the rest of the chain collapses. Then it would appear to be the case that
$$\displaystyle\bigcup_{n=2}^\infty X_n \setminus Y_n = 4\mathbb{N} \setminus 8 \mathbb{N}$$
Is this sound or have I missed something here?
Let $\nu_2(n)$ be the $2$-adic valuation on $\mathbb{N}$. Let $S_k = \{n \in \mathbb{N} : \nu_2(n) \geq k\}$ Your set is:
$(S_2 - S_3) \cup (S_4- S_5) \cup ... = \cup_{k=1}^\infty (S_{2k} - S_{2k+1})$.
$n \in \cup_{k=1}^\infty (S_{2k} - S_{2k+1})$ iff $n \in S_{2k} - S_{2k+1}$ for some $k \geq 1$, iff $\nu_2(n) \geq 2k , \nu_2(n) < 2k+1$ for some $k \geq 1$, iff $\nu_2(n) = 2k$ for some $k \geq 1$.
Therefore, your set is $\{n \in \mathbb{N} : \nu_2(n)$ is even and at least 2 $\}$