In "Modern Algebra" by Seth Warner (1965), section 8 (page 54 in my copy) there exists this:

"If

and

are sets, then there is a bijection from

onto a set

disjoint from

."

It also says: "[This] may be proved in any formal development of the theory of sets."

But I haven't found this proof anywhere, so I'm trying to craft my own proof.

My approach is as follows.

Let

be a bijection, any bijection (the identity mapping will do).

If

then job done,

and

. is our bijection.

Otherwise we have to construct it, or prove that there is a process for doing so.

Take

, and put all elements of

into

.

Let

be the power set of

.

Take some element

.

Then (as is easily proved)

. So put

.

Then you create the mapping

defined as:

Then you've got

and you can repeat the process with

.

However, this gets messy when

is infinite (and worse when it's uncountable) and requires the Axiom of Choice in order to construct such a set.

Also, I haven't proved that the

I picked is not already in

which would render my mapping

not a bijection.

Does anyone know of an easier way of doing this? Apologies for any howlers in this, I'm practically self-taught when it comes to set theory.

Please be aware that any proof produced will appear (in some form) on the page:

Exists Bijection to a Disjoint Set - ProofWiki
... so unless you're happy for this to happen, best not answer.