Show that no two sets in the probability space with $\mathbb{P}(\{k\})=2^{-k!}$ are independent. Let $\mathcal{P}(\mathbb{N})$ denote the power set of $\mathbb{N}$.
Show that no two non-trivial sets in the probability space $(\mathbb{N},\mathcal{P}(\mathbb{N}),\mathbb{P})$ with $\mathbb{P}(\{k\})=2^{-k!}$ for $k\ge2$ and $\mathbb{P}(\{1\})=1-\sum\limits_{k\ge2}\mathbb{P}(\{k\})$ are independent.
There are some hints for this task as follows:
For independent sets $A,B\in\mathcal{P}(\mathbb{N})$, i.e., $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)$,

*

*Show that, wlog $1\notin A\cup B$,

*For $N\ge2$, $\sum\limits_{k>N}2^{N!-k!}\leq\frac{1}{4}$,

*Let $k_A=\min\{k\in A\}$, $k_B=\min\{k\in B\}$, and $k_{AB}=\min\{k\in A\cap B\}$.
Then show that $2^{-k_{AB}!}\leq2\cdot2^{-k_A!-k_B!}$ and $2\cdot2^{-k_{AB}!}\ge2^{-k_A!-k_B!}$. Finally, lead this to a contradiction.

I have shown 2 and 3 but I am unable to show 1 and even after assuming that 1 is true I am stuck after showing 3 and unable to obtain a contradiction.
If someone could help me with these two things I would really appreciate it.
 A: Non-existence of non-constant independent random variables on this space:
Let $X$ and $Y$ be independent random variables on this space and suppose they are both non-constant. Let $E$ be a
non-empty Borel set in $\mathbb{R}$ which does not contain $X(1)$ and $F$ be a non-empty Borel set which does not contain $Y(1)$. We prove that $P\{X^{-1}(E)\cap Y^{-1}(F)\}\neq
P\{X^{-1}(E)\}P\{Y^{-1}(F)\}$. We have $P\{X^{-1}(E)\}=\sum\limits_{X(n)\in
E}^{{}}\frac{1}{2^{n!}},P\{Y^{-1}(F)\}=\sum\limits_{Y(n)\in F}^{{}}\frac{1}{%
2^{n!}}$ and $P\{X^{-1}(E)\cap Y^{-1}(F)\}=\sum\limits_{X(n)\in E,Y(n)\in
F}^{{}}\frac{1}{2^{n!}}$. Let $A=\{n:X(n)\in E\}$ and $B=\{n:Y(n)\in F\}$.
If $P\{X^{-1}(E)\cap Y^{-1}(F)\}=P\{X^{-1}(E)\}P\{Y^{-1}(F)\}$ then we have $%
\sum\limits_{n\in A}^{{}}\frac{1}{2^{n!}}\sum\limits_{n\in B}^{{}}\frac{1}{%
2^{n!}}=\sum\limits_{n\in A\cap B}^{{}}\frac{1}{2^{n!}}.$ This gives $%
\sum\limits_{n\in A,m\in B}^{{}}\frac{1}{2^{n!+m!}}=\sum\limits_{k\in
A\cap B}^{{}}\frac{1}{2^{k!}}$. We look at the two sides as expansions to
base $2$ of some number in $(0,1)$. We note that $n!+m!=k!+j!$ implies $%
(n,m)=(k,j)$ or $(n,m)=(j,k)$. To see this suppose, without loss of
generality, $n$ is the least of the integers $n,m,k,j$ and divide both sides
by $(n+1)!.$ We get $\frac{1}{n+1}\in \mathbb{Z}$
, a contradiction unless $j$ or $k$ equals $n$. If $k=n$ then we get $m!=j!$
so $m=j$. Similarly $j=n$ implies $m=k$. This proves that $(n,m)=(k,j)$ or $%
(n,m)=(j,k)$. Thus in the sum $\sum\limits_{n\in A,m\in B}^{{}}\frac{1}{%
2^{n!+m!}}$ each term is repeated at most twice. If $k\in A\cap B$ we must
have $\frac{1}{2^{k!}}=\frac{1}{2^{n!+m!}}$ or $\frac{1}{2^{k!}}=\frac{2}{%
2^{n!+m!}}$ for some $n$ and $m>1$ [ by uniqueness of expansions to base $2$
]. Hence $n!+m!=k!$ or $n!+m!-1=(k!)$. We note that $n!+m!$ can never be a
factorial ( as can be seen by a very elementary argument), nor can $n!+m!-1$
be a factorial since $-1$ is not divisible by $2!$ Thus $A\cap B$ is empty.
This contradicts the equation $\sum\limits_{n\in A,m\in B}^{{}}\frac{1}{%
2^{n!+m!}}=\sum\limits_{k\in A\cap B}^{{}}\frac{1}{2^{k!}}$ and the proof
is complete.
