Given nonempty sets S and T, does there exist a set R that is disjoint from S with |R|=|T|? Let $S$ and $T$ be nonempty sets. I would like to show that there exists a set $R$ such that $S\cap R=\emptyset$ and $\left\vert T\right\vert = \left\vert R\right\vert$. Here is my work so far. Let $F_S$ and $F_T$ denote the free groups on $S$ and $T$, respectively, and consider the free product $F_S\ast F_T$. Fix $s_0\in S$, and define a mapping 
\begin{align*}
f:T\to F_S\ast F_T
\end{align*}
by 
\begin{align*}
f(t)=s_0t
\end{align*}
for all $t\in T$. Then $f$ is one-to-one, since $f(t)=f(t')\implies s_0t=s_0t' \implies t=t'$. Hence, $f:T\to f(T)$ is a bijection. Now, I'd like to take $R=f(T)$, so I need to show that $S\cap f(T)=\emptyset$. But this seems true: If there exists some $s\in S\cap f(T)$, then $s=s_0t$ for some $t\in T$. But the word length of $s$ is 1, whereas the word length of $s_0t$ is 2; hence, these elements cannot be equal. Is this reasoning correct? And if so, is there a simpler way to establish this result? 
 A: Let $R=T\times\{S\}.$ By folding in $\{S\}$, we've made each element of $R$ "too big" to also be an element of $S$. (One way to make precise this notion of "too big" is the following: the transitive closure of any element of $R$ is strictly larger than the transitive closure of any element of $S$.)
Note that we've used implicitly a non-trivial fact about set theory: that no element of a set $S$ can be of the form $(a, S)$. This is a consequence of the axiom of foundation. In set theories without foundation, this argument doesn't work, and in fact the theorem need not be true: for instance, if there is a "set of all sets" (as in $NF$), then taking that set to be $S$ yields a counterexample.
A: Here is another approach that doesn't appeal to the axiom of foundation. 
It gives a direct construction, for any set $S$, of another set $R$ such that $|S|=|R|$ and $S \cap R= \emptyset$. 
As for your problem, the construction in the link is for the case when $S=T$. 
For all remaining cases of the relationship between $S$ and $T$ the method described below will work.
Use the construction in the link to give you an $R$ for $S \cup T$. Then $R$ being disjoint form $S \cup T$ implies that $R$ is disjoint from all subsets of $S \cup T$; in particular $S$. It also follows that all of the subsets of $R$ are disjoint form $S$. In particular that subset $R' \subset R$ such that $|R'|=|T|$.
We don't need to construct such an $R'$ since only a proof of its existence is required. So we're done.
