I am not sure whether the following construction is recursive or not. Suppose we have an infinite set $X$ and we want to construct a sequence $(x_n)_{n\in \mathbb N}\subset X$ as follows:
- Choose $x_1\in X$
- Let $A_{x_1}\subset X$ such that, $x_1\in A_{x_1}$ and $X\setminus A_{x_1}$ is infinite
- Choose $x_2\in X\setminus A_{x_1}$
- Let $A_{x_1,x_2}\subset X$ such that, $x_2\in A_{x_1,x_2}$ and $X\setminus A_{x_1,x_2}$ is infinite
- Choose $x_3\in X\setminus A_{x_1,x_2}$
- $\ldots$
Can I create a sequence $(x_n)_{n\in \mathbb N}$ in this way? My intuition says yes. It looks like a recursive definition, but I we run the algorithm a second time (even with the same initial value $x_1$), we may end up with a totally different sequence as outcome.
Also, I don't see how we can use the Axiom of Choice to construct a choice function on $\{X, X\setminus A_{x_1}, X\setminus A_{x_1,x_2},\ldots \}$ since all these sets, expect for $X$, depend on all the previous choices, and hence are not pre-defined.
Example: Let $X=\mathbb R$.
- Choose $x_1\in \mathbb R$
- Let $A_{x_1}=\{2^nx_1:n\in \mathbb Z\}$
- Choose $x_2\in \mathbb R\setminus A_{x_1}$
- Let $A_{x_1,x_2}=A_{x_1}\cup \{2^nx_2:n\in \mathbb Z\}$
- Choose $x_3\in \mathbb R\setminus A_{x_1,x_2}$
- $\ldots$
Can we construct a sequence $(x_n)_{n\in \mathbb N}$ in this way? If yes, what theorem can we use to formally guarantee its existence? For instance, can we do the following?
- Choose $x_1\in \mathbb R$
- Assume that we have $x_1,\ldots,x_n$
- Then, choose $x_{n+1}\in X\setminus \bigcup_{i=1}^{n}\{2^nx_i:n\in \mathbb Z\}$
Thank you for your time!