According to Bogachev's book, Ulam's theorem (Theorem 1.12.40) states that if a finite countably additive measure $\mu$ is defined on all subsets of the set $X$ of cardinality $\aleph_1$ and vanishes on all singletons, then it is identically zero. In the proof, by hypothesis, $X$ can be well-ordered in such a way that, for every $y\in X$, the set $\{x:x<y\}$ is at most countable. Then, define an injective function $f$ from this set into $\mathbb{N}$. This means that for every pair $(x,y)$ with $x<y$, $f(x,y)=n$ for some natural number $n$. Bogachev then defined, for every $x\in X$ and every natural $n$, $$A_{x}^{n}=\{y\in X:x<y;f(x,y)=n\}$$
I know that $A_{x}^{n}\cap A_{z}^{n}=\emptyset$ where $x\ne z$. But is $A_{x}^{n}\cap A_{x}^{n'}=\emptyset$ where $n\ne n'$? I'm thinking that it is because if there were a $y\in A_{x}^{n}\cap A_{x}^{n'}$, then $x<y$, $f(x,y)=n$, and $f(x,y)=n'$. That's a contradiction. Am I right?