TITLE: Can any subset of $x$ be moved out of $x$?
QUESTION [8 upvotes]: 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.

REPLY [7 votes]: 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$$(T,x)\in T\Leftrightarrow(T,x)\notin T.$$