Exists Bijection to a Disjoint Set

Theorem
Let $$S$$ and $$T$$ be sets.

Then there exists a bijection from $$T$$ onto a set $$T'$$ disjoint from $$S$$.

Proof
Consider the set:
 * $$X = \left\{{y \in S: \left({\exists x \in T: \left({x, y}\right) \in S}\right)}\right\}$$

That is, $$X$$ consists of all the elements of $$S$$ which are the second coordinate of some ordered pair which also happens to be in $$S$$ and whose first coordinate is in $$T$$.

From Exists Element Not in Set, we have that $$\exists z: z \notin X$$.

Now consider the cartesian product $$T' = T \times \left\{{z}\right\}$$.

Suppose that $$p \in T'$$.

Then $$p = \left({c, z}\right)$$ where $$c \in T$$.

Suppose $$\left({c, z}\right) \in S$$.

Then that would mean that $$z \in X$$.

But we have specifically selected $$z$$ such that $$z \notin X$$.

So $$p = \left({c, z}\right) \notin S$$.

Thus we have that $$T' \cap S = \varnothing$$.

There is an obvious bijection $$g: T \to T'$$:
 * $$\forall t \in T: g \left({t}\right) = \left({t, z}\right)$$

and hence the result.