A decreasing transfinite sequence of subsets of a countable set.

Question

Let $X$ be a countable set and $(S_\alpha)_{\alpha< \rho}$ is a decreasing transfinite sequence of subsets of $X$ in the sense that $$ S_\alpha \supset S_\beta $$ whenever $\alpha<\beta$. Here $\rho$ is some fixed ordinal.

Suppose that $(S_\alpha)_{\alpha< \rho}$ is strictly decreasing, i.e. $S_\alpha \ne S_\beta$ whenever $\alpha<\beta$, how do we show that $\rho$ must be a countable ordinal?

I am sorry if this question is elementary, I have very little training in axiomatic set theory. I think I could prove by contradiction but some crucial steps are missing and I don't know how to make it rigorous.

Answer 1

Here is an alternative proof: Since $X$ is countable, there is a strict well-order $\prec$ of $X$. For each $\alpha < \rho$ let $$ x_\alpha := \min_{\prec} S_\alpha \setminus S_{\alpha + 1}. $$ Now $$ \pi \colon \rho \to X, \alpha \mapsto x_\alpha $$ is an injection, witnessing that $\rho$ is countable (since $X$ is countable).

Answer 2

If it is strictly decreasing, then $A_\alpha=S_\alpha\setminus S_{\alpha+1}$ is non-empty, and $A_\alpha\cap A_\beta=\varnothing$ whenever $\alpha\neq\beta$.

How many pairwise disjoint subsets a countable set can have?