Disjointness in proof of Tarski's Theorem that $|X\times X|=|X|$ for all infinite sets implies Choice

Question

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.

Answer 1

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.

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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 $B=C=X$.