No nontrivial independent sets of events in $2^\mathbb{N}$ From "Theorems and Counterexamples in Mathematics" Exercise 4.1.2
We say that a set $\mathcal{E} = \{A_\lambda\}_{\lambda \in \Lambda}$ of events $A_\lambda$ is independent iff for every subset $\{A_n : A_i \neq A_j \text{if } i\neq j, 1 \leq n \leq N\}$ satisfies $$P\left(\bigcap_{n=1}^N A_n \right) = \prod_{n=1}^N P(A_n).$$ The trivial instance of independence is that for any event $A$, the collection $\{\emptyset,A,X\}$ is independent where $X$ is the whole space.
Let $P$ in the measure situation $(\mathbb{N},2^\mathbb{N},P)$ be the discrete measure $$ P(n) \overset{\text{def}}{=} \begin{cases} 1 - \sum_{n=2}^\infty 2^{-n!} & \text{if } n = 1 \\ 2^{-n!} & \text{if } n \geq 2. \end{cases}$$
I want to show that "trivial instances aside, there are no independent sets of events in $2^\mathbb{N}$". I know that I need to show that for any pair of distinct events $A,B \notin \{\emptyset,\mathbb{N}\}$ we have $P(A \cap B) \neq P(A)P(B)$. There is a first part that I can solve that suggests I should use the fact that if $1 < m,n \in \mathbb{N}$ then there is no $k \in \mathbb{N}$ such that $k! = m! + n!$. However, I can make no sense of how to solve this problem.
 A: As a partial result, I'll resolve the case where $1\not\in A,\,1\not\in B$.

Claim:$\;$If $A,B$ are nonempty subsets of $\mathbb{N}$ with $1\not\in A,\,1\not\in B$, then the events $A,B$ are dependent.

Proof:

Suppose instead that $P(A\cap B)=P(A)P(B)$.

We have $P(A),P(B) > 0$, hence $P(A\cap B) > 0$, so
$A\cap B\ne{\large{\varnothing}}$.

Also we have $P(A\cap B) < P(A)$ and $P(A\cap B) < P(B)$, hence neither of $A,B$ is a subset of the other.

Let $a=\min(A),\,b=\min(B),\,c=\min(A\cap B)$.

Without loss of generality we can assume $a\le b$.

Then since $B$ is not a subset of $A$, it follows that $A$ is proper subset of $\{a,a+1,a+2,...\}$

Then we get
\begin{align*}
2^{-a!}\le &P(A) < \sum_{k=a!}^\infty 2^{-k}=2^{-a!+1}\\[4pt]
2^{-b!}\le &P(B)\le\sum_{k=b!}^\infty 2^{-k}=2^{-b!+1}\\[4pt]
2^{-c!}\le &P(A\cap B)\le\sum_{k=c!}^\infty 2^{-k}=2^{-c!+1}\\[4pt]
\end{align*}
which implies
\begin{align*}
2^{-a!-b!}\le &P(A)P(B) < 2^{-a!-b!+2}\\[4pt]
2^{-c!}\le &P(A\cap B)\le 2^{-c!+1}\\[4pt]
\end{align*}
and hence
\begin{align*}
-a!-b!&\le -c!+1\\[4pt]
-c! &< -a!-b!+2\\[4pt]
\end{align*}
and then since $a!,b!,c!$ are even, we get
$$
-a!-b!\le -c!\le -a!-b!
$$
so $a!+b!=c!$.

But we have
$$
a! < a!+b!\le 2a! < (a+1)!
$$
so we get $a! < c! < (a+1)!$, contradiction.
