Disjointness in proof of Tarski's Theorem that $|X\times X|=|X|$ for all infinite sets implies Choice In the Wikipedia proof of Tarski's Theorem about Choice, one assumes that the infinite set $B$ and the ordinal $\beta$ are disjoint.
While it is clear that this assumption can be made, is it actually necessary in the rest of the proof? I cannot locate where this assumption is used.
 A: No, it's not necessary. But it makes it a bit easier when we talk about $\gamma\in\beta$ and $\gamma\notin B$. It is also a bit simpler conceptually to separate the two sets instead of dealing with their intersection.
Sometimes adding a small and unnecessary assumption can clarify the concept behind the proof, and when the unnecessary assumption does not hurt the generality, this is even more welcome.

Let me remark that it is unfortunate that the proof is presented the way that it is. Instead the correct way is to extract the following lemma:

Lemma 1. Suppose that $X$ is a set and $\lambda$ is an ordinal. If $|X\cup\lambda|=|X\times\lambda|$, then $|X|$ and $\lambda$ are comparable.

This, in turn, can be improved and generalised into,

Lemma 2. Suppose that $f\colon C\cup D\to A\times B$ is a surjection, then $C$ maps onto $A$ or $D$ maps onto $B$.

Although it is in fact enough to consider the injective version of this lemma,

Lemma 3. Suppose that $f\colon A\times B\to C\cup D$ is injective, then $A$ injects into $C$ or $B$ injects into $D$.

It is not hard to see why Lemma 2 implies Lemma 3, and Lemma 3 implies Lemma 1 by taking $A=D=\lambda$ and $B=C=X$.
