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Post Closed as "Not suitable for this site" by Emil Jeřábek, Dave Benson, Sam Hopkins, LSpice, bof
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LSpice
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proof Proof of the axiom of choice for finite sets in ZF

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Sam Hopkins
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Let the set $A$ be finite and $\emptyset \notin A$. How can I, without using the axiom of choice, prove by mathematical induction that there exists a function $f : A \rightarrow \bigcup A$, i.e., satisfying $f(a) \in a$ for all $a \in A$?

Let the set $A$ be finite and $\emptyset \notin A$. How can I, without using the axiom of choice, prove by mathematical induction that there exists a function $f : A \rightarrow \bigcup A$, i.e., $f(a) \in a$ for all $a \in A$?

Let the set $A$ be finite and $\emptyset \notin A$. How can I, without using the axiom of choice, prove by mathematical induction that there exists a function $f : A \rightarrow \bigcup A$ satisfying $f(a) \in a$ for all $a \in A$?

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proof of the axiom of choice for countablefinite sets in ZF

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