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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Tarski's theorem about choice ： ウィキペディア英語版 Tarski's theorem about choice In mathematics, the Tarski's theorem, proved by , states that in ZF the theorem "For every infinite set $A$, there is a bijective map between the sets $A$ and $A\backslash times\; A$" implies the axiom of choice. The opposite direction was already known, thus the theorem and axiom of choice are equivalent. Tarski told 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 $A$: $|A|=|A\backslash times\; A|$". 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, $\backslash beta$, such that there is no surjective function from $B$ to $\backslash beta$. We assume without loss of generality that the sets $B$ and $\backslash beta$ are disjoint. By our initial assumption, $|B\; \backslash cup\; \backslash beta|=|(B\; \backslash cup\; \backslash beta)\; \backslash times\; (B\; \backslash cup\; \backslash beta)|$, thus there exists a bijection $f:\; B\; \backslash cup\; \backslash beta\; \backslash to\; (B\; \backslash cup\; \backslash beta)\; \backslash times\; (B\; \backslash cup\; \backslash beta)$. For every $x\; \backslash in\; B$, it is impossible that $\backslash beta\; \backslash times\; \backslash \; \backslash subseteq\; f()$, because otherwise we could define a surjective function from $B$ to $\backslash beta$. Therefore, there exists at least one ordinal $\backslash gamma\; \backslash in\; \backslash beta$, such that $f(\backslash gamma)\; \backslash in\; \backslash beta\; \backslash times\; \backslash$, thus the set $S\_x=\backslash$ is not empty. With this fact in our mind we can define a new function: $g(x)=\backslash min\; S\_x$. This function is well defined since $S\_x$ is a non-empty set of ordinals, hence it has a minimum. Recall that for every $x,y\; \backslash in\; B,\; x\; \backslash neq\; y$ the sets $S\_x$ and $S\_y$ are disjoint. Therefore, we can define a well order on $B$, for every $x,\; y\; \backslash in\; B$ we shall define $x\; \backslash leq\; y\; \backslash iff\; g(x)\; \backslash leq\; g(y)$, since the image of $g$, i.e. $g()$, is a set of ordinals and therefore well ordered. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Tarski's theorem about choice」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.