Can any subset of $x$ be moved out of $x$? Let $x$ be a set and let $y\subset x$. Does there exist a set $z$ such that:
(1) $z\cap x=\emptyset$ and  
(2) there exists a bijection $y \to z$ ? 
It is quite intuitive that the answer should be yes. My first attempt was to take a set $x'\notin x$ and to consider $z=y\times \{ x'\}$. But I am unable to show $z\cap x=\emptyset$. I could imagine a proof with the axiom of choice, but I'd prefer to avoid it if possible.
 A: Edit: simplified version
The Axiom of Foundation is not needed.
Theorem. Given a set $X$ we can find a set $Y$, disjoint from $X$, and a bijection from $X$ to $Y$.
Proof. Let $$T=\{(S,x):(S,x)\in X\wedge(S,x)\notin S\}$$and let$$Y=\{(T,x):x\in X\}.$$Of course $x\mapsto (T,x)$ is a bijection from $X$ to $Y$. Assume for a contradiction that $X\cap Y\ne\emptyset$. This means that there is an element $x\in X$ such that $(T,x)\in X$. Now, according to the definition of $T$,$$(T,x)\in T\Leftrightarrow(T,x)\in X\wedge(T,x)\notin T.$$Since $(T,x)\in X$ by assumption, we arrive at the contradiction$$(T,x)\in T\Leftrightarrow(T,x)\notin T.$$
A: Choosing $z = y\times \{x\} = \bigl\{ \{\{a\},\{a,x\}\} : a \in y\bigr\}$ works, since the axiom of foundation guarantees that $z \cap x = \varnothing$ then. For otherwise, there'd be an $a\in y$ with
$$x \in \{a,x\} \in \{\{a\},\{a,x\}\} \in x.$$
A: To elaborate on Daniel Fischer's comment, we can prove this using the axiom of foundation, or more easily (in my opinion) using the consequence that every set has an ordinal rank.
Take your set $x'$ to be any set with rank at least that of $x$, say $x' = x$ itself.  Then every element of $y \times \{x'\}$ will have rank greater than that of $x$ because the rank of an ordered pair is greater than the ranks of its components.  On the other hand, every element of $x$ has rank less than that of $x$.  Therefore $x \cap (y \times \{x'\}) = \emptyset$.
