This is a follow-up to the question on field reductions.
(EDIT: Originally this question used the notation $\{\mathbb{R}(\setminus a)\}$, but now uses $\mathbb{R}(\setminus a)$ instead.)
There isn't a unique largest subfield of $\mathbb{R}$ not containing $a$, so let $\mathbb{R}(\setminus a)$ denote the set of maximal subfields of $\mathbb{R}$ not containing $a$,
EDIT: The question "What is the cardinality of $\mathbb{R}(\setminus a)$ ?" has now been asked on Mathoverflow.
From the previous question if $F\in \mathbb{R}(\setminus a)$ then $F$ is uncountable, but what is the size of $\mathbb{R}\setminus F$ the complement of $F$ in $\mathbb{R}$
Each $F \in \mathbb{R}(\setminus a)$ should be arrived at by removing elements of $\mathbb{R}$, but if you start with $\mathbb{R}$ and remove $a$ and then remove more elements until you get a field then surely you arrive at a single field, so where do the multiple maximal fields come from ? - It must be because you get different fields depending on the order in which elements are removed from $\mathbb{R}$, in which case what is the most natural order to remove elements to get a definition of a single canonical subfield $\mathbb{R}(\setminus a)$ of $\mathbb{R}$ that doesn't contain $a$, that we could call 'the' field reduction of $\mathbb{R}$ by $a$ ?
EDIT: Given Arturo Magidin's comments and answer, if we abandon the idea of a distinguished element in the abstract/in general, then since the notation $\{\mathbb{R}(\setminus a)\}$ looks like a set containing a single element called $\mathbb{R}(\setminus a)$, it makes more sense to drop the brackets and just write $\mathbb{R}(\setminus a)$ on the understanding that this is a set of field reductions i.e. a set of subfields not a single subfield.
From Pete L. Clark's answer to the previous question, if $F(a)=\mathbb{R}$ then $F=\mathbb{R}$, so let's define the notion of a super-extension of $F$ by $a$, $F(a..)=R$ being the sequence of field extensions to get to $\mathbb{R}$ from $F\in\mathbb{R}(\setminus a)$. More generally, let $@K(\setminus a)$ be an element of $K(\setminus a)$ then $@K(\setminus a)(a..)=K$.
We could also reduce $K$ by a set $A$, with $K(\setminus A)$ being the set of maximal subfields that don't contain any elements from $A$, and again write the super-extension $@K(\setminus A)(A..)=K$.
Let $\mathbb{A}'$ be the set of non-rational algebraic numbers.
What is the cardinality of $\mathbb{R}(\setminus \mathbb{A}')$ ?