I am currently studying Real Analysis by Folland and in a section on well-ordering he gives the following proof:
Proof of existence of countable set of ordinals
Essentially the proof proceeds as follows: We want to find an uncountable, well-ordered set $Q$ such that for each $q\in Q$, $I_q$ is countable. Here $I_q = \{p \in Q \mid p < q\}$ (i.e. the predecessors of $q$ in $Q$ wrt the well-ordering on $Q$)
- There exist uncountable well-ordered sets by the well-ordering principle
- Choose such a set, $X$. If $X$ has the property we are done.
- Otherwise, there is a minimal $x_0$ such that $I_{x_0}$ is uncountable, in which case $Q = I_{x_0}$
I am confused by step 3, if the set of predecessors of $x_0$ is uncountable how can the set of predecessors for any element of this set be countable? If I choose any element in $I_{x_0}$, it should have uncountably many predecessors right? Hence this set doesn't actually have the property we want?
I'm also quite confused at what it means to find a minimal element of an uncountable set. For instance if we consider the reals, $\mathbb R$, this is an uncountable set with a natural well-ordering, but it does not seem obvious to me that the notion of a "minimal" element makes any sense here, especially given that for all $x$ in $\mathbb R$, $I_x$ should be uncountable. (My logic here is that one can show an injection from $[0,1]$ to $\mathbb R$, hence any other bounded range I choose can be mapped to $\mathbb R$ as well and is therefore uncountable).