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.