Theorem: Assuming the Axiom of Countable Choice, an infinite set $A$ is Dedekind-infinite.

Lemma 1: assuming the Axiom of Countable Choice, if $A$ is an infinite set, then $A$ contains a countably infinite subset.

Please check my below proof! Thank you for your help!

By Lemma 1, there exists a countably infinite subset $B$ of $A$. Let $(b_1,b_2,\cdots)$ be a listing of the elements of $B$, without repetition. We define a function $f:A \to A$ by $f(b_i)=b_{2i}$ for all $i \in \mathbb N$ and $f(a)=a$ for all $a \in A \setminus B$. Then $f$ is clearly injective. Furthermore, $f(A)=A \setminus \{b_1,b_3,b_5,\cdots\}$, then $f(A) \subsetneq A$. Hence $f:A \to f(A)$ is a bijection from $A$ to a proper subset of $A$. Consequently, $A$ is Dedekind-infinite.

  • $\begingroup$ looks right to me. You could have also chosen $f(b_i) = b_{i+1}.$ $\endgroup$ – spaceisdarkgreen Aug 31 '18 at 4:28
  • $\begingroup$ @spaceisdarkgreen Oh that's much simpler ^^ As a result, $f(A)=A \setminus \{b_1\}$. $\endgroup$ – Le Anh Dung Aug 31 '18 at 4:51

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