Is every infinite set equipotent to a field? For example, $\mathbb N$ is equipotent to $\mathbb Q$ which is a field.
$\mathbb R$ is equipotent to itself, which is a field.
But what about $\mathbb R^{\mathbb R}$, $P(\mathbb R^{\mathbb R})$ etc.?
 A: Is it true that for any infinite set $E$, the cardinal of the field $\mathbb Q(E)$ is equal to the cardinal of $E$?  
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What is $\mathbb Q(E)$?
Here $E$ is considered to be a set of "indeterminates".  Or, if that is not convenient, invent a set of indeterminates perhaps with some notation like $X_e$, one for each element $e \in E$.  Then $\mathbb Q[E]$ is the set of polynomials in these indeterminates, with coefficients in $\mathbb Q$.  (Of course each polynomial involves only finitely many of the elements of $E$.)  
So $\mathbb Q(E)$ is a ring.  It is an algebra over $\mathbb Q$.  It is an integral domain.
And $\mathbb Q(E)$ is the set of rational functions in these indeterminates.  In other words, formulas of the type $f/g$, where $f,g \in \mathbb Q[E]$ and $g \ne 0$.  
So $\mathbb Q(E)$ is a field.  It has the set $E$ (or a set $\{X_e : e \in E\}$ identified with $E$) of mutually algebraically independent elements that generate it (as a field over $\mathbb Q$).  
To compute the cardinality of $\mathbb Q(E)$, compute in turn: how many monomials in $E$ are there; how many linear combinations of monomials (i.e. polynomials); how many quotients of those (rational functions).
A: The field axioms are a countable set over a countable language. Since there is an infinite model of the field axioms, there is (by Lowenheim-Skolem) a model of every infinite cardinality.
A: For every infinite set $X$ there is a field $\mathbb F$ which is equipotent to $X$. In fact, much more is true. This is an immediate consequence of the Upward Löwenheim-Skolem Theorem. Fix a countable field $K$ (e.g. $\mathbb Q$). By the Löwenheim-Skolem Theorem there is some $\mathbb F$ such that $\mathbb F$ is equipotent to $X$ and $K \prec \mathbb F$. This implies that $K$ and $\mathbb F$ have the same theory and since $K$ is a field, $\mathbb F$ is a field as well.
A: I think this answer by Gregory Grant is in some ways better than any relying on the Löwenheim–Skolem theorem, since it gives an explicit construction of a field of arbitrary infinite cardinality.
