A while ago I needed (a lemma in a larger proof) to construct for any arbitrary set A, another set that is disjoint from A and of the same cardinality as A (without the use of higher machinery of ordinals). The task is harder than it may seem. One may choose to attempt a proof by contradiction. One of my friends has in fact proceeded that way. I came up with the following construction which I’m quite fond of.

First, let me state a theorem (which is a direct consequence of Russell’s paradox) that I helped myself to.

**Theorem**. For each set S, the set

Is not a member of S.

For if we assume otherwise, an immediate contradiction ensues, for

Now for the construction. Let’s take some set A, and construct a set B that is disjoint from it, and equinumerous with it. Consider the set

**Theorem**: For any set A the following set B is disjoint from A and equinumerous with it:

Less formally

It is clear that B has the same cardinality as A. (post-Clarification: just to be pedantic, it's not just the triple x, y, z that is in A, but rather I should have written "for x, y, z,... in A". I forgot the ellipsis.)

It is also clear that,

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