Non-existence of independent sets

Question

I found the following exercise quite intriguing yet had little success getting on the right track. Suppose $(\Omega, \mathcal{F}, \mathbb{P}) = (\mathbb{N}, \mathcal{P}(\mathbb{N}), \mathbb{P})$ with $\mathbb{P}(\{n\}) = (\frac{1}{2^{n!}})$ for all $n \geq 2$ and $\mathbb{P}(\{1\}) = 1 - \sum\limits_{n = 2}^{\infty} \mathbb{P}(\{n\})$. Prove that there are no nontrivial sets $F_1, F_2 \in \mathcal{P}(\mathbb{N})$ that are independent. I assumed the opposite for contradiction but arrived nowhere. What is the way to go here?

Answer 1

We cannot have two non-constant independent random variables on this space. Let $X$ and $Y$ be independent random variables on $(\Omega ,\mathcal F ,P)$ 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 $$P\{X^{-1}(E)\}=\sum_{X(n)\in E}\frac{1}{2^{n!}},P\{Y^{-1}(F)\}=\sum_{Y(n)\in F}^{{}}\frac{1}{% 2^{n!}}$$ and $$P\{X^{-1}(E)\cap Y^{-1}(F)\}=\sum_{X(n)\in E,Y(n)\in F}^{{}}\frac{1}{2^{n!}}$$. 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 $$\sum_{n\in A}^{{}}\frac{1}{2^{n!}}\sum_{n\in B}^{{}}\frac{1}{ 2^{n!}}=\sum_{n\in A\cap B}^{{}}\frac{1}{2^{n!}}$$ This gives $$ \sum_{n\in A,m\in B}^{{}}\frac{1}{2^{n!+m!}}=\sum_{k\in A\cap B}^{{}}\frac{1}{2^{k!}}$$. 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 $\frac{1}{n+1}\in \mathbb N$ 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 $\sum_{n\in A,m\in B}^{{}}\frac{1}{% 2^{n!+m!}}$ 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 $\sum_{n\in A,m\in B}^{{}}\frac{1}{% 2^{n!+m!}}=\sum_{k\in A\cap B}^{{}}\frac{1}{2^{k!}}$ and the proof is complete.