Proposition: let $A$ and $B$ be arbitrary sets. Then, there exists a set $C$ such that $A\cap C=\varnothing$ and $C\sim B$.
This proposition seems really easy at first, but when I tried to prove it, it got me really thinking a lot about it. In the end, I think I managed to prove it succesfully, though I used the axiom of choice in order to do so. I write here my idea together with a sketch of my proof:
The first set that one knows that does not belong to $A$ is $A$ itself, so if $B$ had cardinality $1$, then $C=\{A\}$ would work. Similarly, if $B$ had cardinality $2$, any other set $X$ such that $A\in X\neq\{A\}$ would give a working set $C=\{A,X\}$ (this is using axiom of regularity). I think after having worked a bit with ordinals that it is natural to think $X=S(A)=A\cup\{A\}$. Using the replacement axiom, it is posible to define a recursive sequence whose image is $\{A,S(A),S(S(A)),...\}$ and if $B$ were numerable, then $C=\{A,S(A),S(S(A)),...\}$ would work. However, this led me to see how I could do this in general for any ordinal, but this will use that $B$ is equipotent to some ordinal.
The idea is then to have a copy of the ordinals, starting on $A$ rather than on the empty set.
Define a transfinite sequence as follows:
- $x_0=A$.
- For every ordinal $\alpha$, $x_{\alpha+1}=S(x_\alpha)$.
- For every limit ordinal $\lambda$, $x_\lambda=\bigcup_{\delta<\lambda}x_\delta$.
The following facts are then true (I won't include proof for this but many of the arguments are by transfinite induction):
- For any two ordinals $\alpha,\beta$, if $\alpha<\beta$ then $x_\alpha\in x_\beta$.
- For any ordinal $\alpha$, $\alpha\sim C_\alpha:=\{x_\delta|\delta<\alpha\}$ (which exists by axiom of replacement).
- For every non-zero ordinal $\alpha$, $A\in x_\alpha$.
- For any ordinal $\alpha$, $A\cap C_\alpha=\varnothing$.
With these facts on hand, it is possible to give now the argument. Using the axiom of choice, $B$ can be well ordered, and so it must be equipotent to some ordinal $\gamma_B$. Now, we have $B\sim\gamma_B\sim C_{\gamma_B}$, and by transitivity, $C_{\gamma_B}$ satisfies $B\sim C_{\gamma_B}$. Also, $A\cap C_{\gamma_B}=\varnothing$ and therefore $C=C_{\gamma_B}$ is a set that satisfies the required conditions. End of the proof.
For this, I required quite some notions of ordinals and axiom of choice, however it seems as a rather elementary proposition, so now I ask: is there an easier way to prove this, using more basic tools from set theory? And also, independently from simpleness, is there a way to prove this wothout using the axiom of choice?
Thanks for reading and any reply is welcome :).
Greetings.