Tarski's theorem about choice

In mathematics, the Tarski's theorem, proved by Alfred Tarski (1924), states that in ZF the theorem "For every infinite set $A$, there is a bijective map between the sets $A$ and $A\times A$" implies the axiom of choice. The opposite direction was already known, thus the theorem and axiom of choice are equivalent.

Tarski told Jan Mycielski (2006) that when he tried to publish the theorem in Comptes Rendus de l'Académie des Sciences Paris, Fréchet and Lebesgue refused to present it. Fréchet wrote that an implication between two well known propositions is not a new result. Lebesgue wrote that an implication between two false propositions is of no interest.

Proof

Our goal is to prove that the axiom of choice is implied by the statement "For every infinite set $A$: $|A|=|A\times A|$". It is known that the well-ordering theorem is equivalent to the axiom of choice, thus it is enough to show that the statement implies that for every set $B$ there exist a well-order.

For finite sets it is trivial, thus we will assume that $B$ is infinite.

Since the collection of all ordinals such that there exist a surjective function from $B$ to the ordinal is a set, there exist a minimal non-zero ordinal, $\beta$, such that there is no surjective function from $B$ to $\beta$. We assume without loss of generality that the sets $B$ and $\beta$ are disjoint. By our initial assumption, $|B \cup \beta|=|(B \cup \beta) \times (B \cup \beta)|$, thus there exists a bijection $f: B \cup \beta \to (B \cup \beta) \times (B \cup \beta)$.

For every $x \in B$, it is impossible that $\beta \times \{x\} \subseteq f[B]$, because otherwise we could define a surjective function from $B$ to $\beta$. Therefore, there exists at least one ordinal $\gamma \in \beta$, such that $f(\gamma) \in \beta \times \{x\}$, thus the set $S_x=\{\gamma | f(\gamma) \in \beta \times \{x\}\}$ is not empty.

With this fact in our mind we can define a new function: $g(x)=\min S_x$. This function is well defined since $S_x$ is a non-empty set of ordinals, hence it has a minimum. Recall that for every $x,y \in B, x \neq y$ the sets $S_x$ and $S_y$ are disjoint. Therefore, we can define a well order on $B$, for every $x, y \in B$ we shall define $x \leq y \iff g(x) \leq g(y)$, since the image of $g$, i.e. $g[B]$, is a set of ordinals and therefore well ordered.

References

• fr:Ordinal de Hartogs#Produit cardinal