This question is from Hrbacek and Jech's intro set theory book (Ch 10, exercise 4.6). I am given:
$$F=\{1\}\cup\{1-\frac{1}{2^{n_1}}-\frac{1}{2^{n_1+n_2}}-...-\frac{1}{2^{n_1+...+n_k}} : k\leq n_1 \text{ and }n_i \geq 1 \text{ for all } i\leq k\}.$$
The notation here seems sort of vague to me, but I am interpreting the $k$, $n_1$ and $n_i$ to be variables in the naturals in particular (so $k$ is not a fixed constant, it is being quantified over in the definition of F). Please correct me if this is wrong.
The actual exercise is to show $F$ satisfies $F_\omega = \{1\}$ where $F_\omega = \cap_{n\in\mathbb{N}} F_n$ for $F_n$ the nth derived set of $F$.
For what I have tried so far, I have let:
$$A_0=\{1\}$$and for each natural number $m$$$A_m = \{1-\frac{1}{2^{n_1}}-\frac{1}{2^{n_1+n_2}}-...-\frac{1}{2^{n_1+...+n_m}} : m\leq n_1 \text{ and }n_i \geq 1 \text{ for all } i\leq k\}$$
So now for $A_m$ the value $m$ is a fixed rather than quantified over.
From this I have defined:
$$B_m = \bigcup_{k=0}^m A_k.$$
It should follow:
$$F=\bigcup_{n=0}^\infty B_m.$$
Using work from a previous exercise I'm pretty sure I can show $B_m \subseteq (B_{m+1})'$, for each choice of $m$ and from $B_n \subseteq F$ it should follow by taking derived sets $n$ times on both sides that $\{1\}\subseteq F_n$, thus $\{1\}\subseteq F_\omega$.
I know next I need to show every element of $F_\omega$ is $1$. So I have tried letting $y$, be an element of $F_\omega$, but I have no idea why $y$ can't be some value other than $1$?