So we were reading Lang's Book of algebra. In chapter V, p. 231 he tries to construct an extension of a field $k$ in which a polynomial has a root. There is a part where Lang says the following:
Let $S$ be a set whose cardinality is the same as that of $F-\sigma k$ ($=$ the complement of $\sigma k$ in $F$) and which is disjoint from $k$.
In other words he claims the following: Given two sets $A$ and $B$, there is a set $B'$ such that $B$ and $B'$ have the same cardinality and $A$ and $B'$ are disjoint.
Someone pointed out if that was always possible, I immediately thought yes, but how is this done exactly? The first thing that came to my mind was to consider something like $\{0\}\times A$ and $\{1\}\times B$, however we end up with a different $A$. I tried modifications of this idea but nothing worked.
I know this really isn't relevant to the theory because one can consider injective homomorphism of fields instead of subfields, but it still caught my attention.
So at the end the question is how do we prove that, again, given two sets $A$ and $B$, there is a set $B'$ such that $B$ and $B'$ have the same cardinality and $A$ and $B'$ are disjoint.
This must be closely related to the foundations of mathematics of which I don't know much, so comments on that are welcome.
If you could find a set $X$ such that no ordered pair in $A$ has the form $(X,y)$, then $\{X\}\times B$ is going to be disjoint from $A$. That's just how things work.
But now that's easy. Since there is no universal set (assuming standard set theory, e.g. ZF), we can engineer such $X$ quite easily: for example, $X=\{x\in\operatorname{dom} A\mid x\notin x\}$ satisfies this property.
If you want to use the axiom of regularity, then even taking something as $A$ itself will work. But that is not necessary, since even without assuming regularity holds, we can prove there is no universal set.