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  "body": "\n## Proof\n\nThe 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|0N109NG5ZX]] is equivalent to the axiom of choice; thus it is enough to show that the statement implies that for every set $B$ there exists a [[well-order|01CGSJ0GNQ]].\n\nSince the collection of all [[ordinals|0RDJK7EQYY]] such that there exists a [[surjective function|01RVCMSN5N]] from $B$ to the ordinal is a set, there exists an infinite ordinal, $\\beta,$ such that there is no [[surjective function|01RVCMSN5N]] from $B$ to $\\beta.$ We assume without loss of generality that the sets $B$ and $\\beta$ are [[disjoint|0W4RPA8QH3]]. By the initial assumption, $|B \\cup \\beta| = |(B \\cup \\beta) \\times (B \\cup \\beta)|,$ thus there exists a [[bijection|0GG9TVW90M]] $f : B \\cup \\beta \\to (B \\cup \\beta) \\times (B \\cup \\beta).$\n\nFor 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 $f(\\gamma) \\in \\beta \\times \\{x\\},$ so the set $S_x = \\{\\gamma : f(\\gamma) \\in \\beta \\times \\{x\\}\\}$ is not empty.\n\nWe 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, and so has a minimum. 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 define $x \\leq y \\iff g(x) \\leq g(y),$ since the image of $g,$ that is, $g[B]$ is a set of ordinals and therefore well ordered.\n",
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