Field reductions. part two 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 $\mathbb{R}(\setminus \mathbb{A}')$ ?
 A: I'll place something as an answer in the interest of not having questions with no answers.
I don't know what the cardinality of $\{\mathbb{R}(\setminus a)\}$ is, but I would be surprised if it is not uncountable. I'll see if I can think about it some in the next couple of days.
As for your final paragraph, the different fields come not from the order in which you would remove elements, but rather on the choices you make as to what to remove at each (of infinitely many) steps. As I noted in the comments, think of the similar process whereby we look for subgroups of a given group that are maximal among those that do not contain a given element. If we start with the Klein $4$-group, $C_2\times C_2$, and remove the element $a=(1,1)$, then we get two different choices of what other element to remove in order to get a group: we can remove either $(1,0)$ 
or 
$(0,1)$; each of the choices will lead to a different subgroup. 
Similarly, consider for instance the case where $a=\sqrt{2}$; then as noted in your previous question, a maximal subfield of $\mathbb{R}$ that does not contain $a$ may contain either $\sqrt{3}$ or $\sqrt{6}$, but not both, and you get one field that is maximal and does contain $\sqrt{3}$ (simple application of Zorn's Lemma), and one that is maxiimal and does contain $\sqrt{6}$ (ditto); so with your process you can either remove $\sqrt{3}$ at some point, or remove $\sqrt{6}$ at some point, and these choices lead to different maximal subfields.
The very word "choices" suggests how the Axiom of Choice is playing a role in this "top-down" construction of elements of $\{\mathbb{R}(\setminus a)\}$ (in the "bottoms-up" approach, it plays a role via Zorn's Lemma). In order to be able to obtain something one could call the "field reduction", we would need some procedure or algorithm that determines which choice we make at any stage. One way would be by endowing $\mathbb{R}$ with a well-ordering, and then at each step we can consider the set of all elements we can remove; if it is empty, we are done; if it is not empty, we remove the least element of the set. But, of course, well-ordering $\mathbb{R}$ is a bit of a problem in and of itself: there is no consistently agreed-upon well-ordering of $\mathbb{R}$ (and in some cases, not even an explicit one). We could try to pick a particular model in which there are explicit well-orderings (as mentioned by Adres Caicedo in the comments, such models exist), and a particular well-ordering in there, and then define "the" field reduction, but that seems to be as much of a problem in and of itself: which model, and which order, and why? Perhaps there are other mechanisms whereby one could try to establish a distinguished element in $\{\mathbb{R}(\setminus a)\}$ (or perhaps, a distinguished element for specific instances of $a$), but it seems to me to be a hopeless task in the abstract/in general. 

Added and edited: Recall that any field extension $K/k$ can be broken up as $K/F/k$, where $K/F$ is algebraic and $F/k$ is a purely transcendental extension. In particular, if $T$ is a transcendence basis for $K$ over $k$, then $k(T)$ will be a maximal subfield of $K$ that contains $k$ and does not contain any elements of $K$ that are algebraic over $k$ but not in $k$. So for any field extension $K/k$, if we let
\begin{equation*}
\mathbb{A}_{K/k} =\bigl\{ a \in K\ \bigm|\ \mbox{$a$ is algebraic over $k$, $a\notin k$}\bigr\},
\end{equation*}
then 
$$K(\setminus \mathbb{A}_{K/k}) = \{ k(T)\ |\ \mbox{$T$ is a transcendence basis for $K$ over $k$}\}.$$
If $K$ is algebraic over $k$, then you just get $k$, of course; if $K$ is purely transcendental over $k$, then $\mathbb{A}_{K/k}$ is empty, so you get $K$. 

I had originally written that the set would always be singleton, but this is not the case. If $T$ is any transcendence basis for $\mathbb{R}$ over $\mathbb{Q}$, you can modify an element of $T$ by multiplying it by a non-rational algebraic number and you get a slightly different subfield. For example, fix $t\in T$ and replace $t$ with $\sqrt{2}t$ to obtain a new set $T'$. This is still a transcendence basis for $\mathbb{R}$ over $\mathbb{Q}$; but whereas $\sqrt{2}t\in \mathbb{Q}(T')$, it does not lie in $\mathbb{Q}(T)$. If it did, then we would immediately get $\sqrt{2}\in\mathbb{Q}(T)$, contradicting the fact that $\mathbb{Q}(T)$ is purely transcendental over $\mathbb{Q}$. You can do the same thing taking arbitrary subsets of $T$ and multiplying them through by a nonrational algebraic number. Since $|T|=\mathfrak{c}$, this gives at least $2^{\mathfrak{c}}$ distinct elements in $\mathbb{R}(\setminus\mathbb{A}')$, and since that is the number of subsets of $\mathbb{R}$, it follows that the cardinality of $\mathbb{R}(\setminus\mathbb{A}')$ is $2^\mathfrak{c}$. 
