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Version 4 |
| Following is a proof that the axiom of choice implies Zermelo's postulate. |
Following is a proof that the axiom of choice implies Zermelo's postulate. |
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| \begin{proof} |
\begin{proof} |
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Let $\mathcal{F}$ be a disjoint family of nonempty sets. Let $\displaystyle f \colon \mathcal{F} \to \bigcup \mathcal{F}$ be a choice function. Let $A,B \in \mathcal{F}$ with $A \neq B$. Since $\mathcal{F}$ is a disjoint family of sets, $\displaystyle A \cap B = \emptyset$. Since $f$ is a choice function, $f(A) \in A$ and $f(B) \in B$. Thus, $f(A) \notin B$. Hence, $f(A) \neq f(B)$. It follows that $f$ is injective.
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Let $\mathcal{F}$ be a disjoint family of nonempty sets. Let $\displaystyle f \colon \mathcal{F} \to \bigcup \mathcal{F}$ be a choice function. Let $A,B \in \mathcal{F}$. Since $\mathcal{F}$ is a disjoint family of sets $\displaystyle A \cap B = \emptyset$. Since $f$ is a choice function, $f(A) \in A$ and $f(B) \in B$. Thus, $f(A) \notin B$. Hence, $f(A) \neq f(B)$. It follows that $f$ is injective.
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| Let $\displaystyle C=\left\{f(B) \in \bigcup \mathcal{F} : B \in \mathcal{F} \right\}$. Then $C$ is a set. |
Let $\displaystyle C=\left\{f(B) \in \bigcup \mathcal{F} : B \in \mathcal{F} \right\}$. Then $C$ is a set. |
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| Let $A \in \mathcal{F}$. Since $f$ is injective, $\displaystyle A \cap C=\{f(A)\}$. |
Let $A \in \mathcal{F}$. Since $f$ is injective, $\displaystyle A \cap C=\{f(A)\}$. |
| \end{proof} |
\end{proof} |
