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I am finding this proof to be difficult and would appreciate some feedback on my current approach.

EDIT: I've since revised my solution. My first solution claimed a map was injective when in fact it was not. Thanks everyone.

Euler's Friend
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    Your $g$ is not injectiive as $g(a_1)=b_1=g(b_1)$. You should have suspected something's wrong because your argument would work also for finite $S$ if only $|S|\ge 2|A|$. – Hagen von Eitzen Jul 20 '19 at 19:23
  • @HagenvonEitzen Funny, I saw the flaw as soon as I posted. I will continue trying things. Thank you. – Euler's Friend Jul 20 '19 at 19:26
  • So you must use the fact that $S$ is infinite. Maybe you can consider picking up an countably infinite sequence $S' \subseteq S$ (which must exist) and do something on it... – Hw Chu Jul 20 '19 at 19:45
  • Related: https://math.stackexchange.com/q/2915402/524154 –  Jul 20 '19 at 23:02

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As hinted by Hw Chu, you can construct an infinite countable set $A'\subset S$ such that $A\subset A'$ then you can construct a sequence $ (a_{n})_{n\in \mathbb{N^{*}}} $ with obvisouly $ A=\{a_{1},…,a_{n}\} $ i.e A is the first n elements of the sequence and $ \{a_{n}\mid n\in \mathbb{N^{*}} \}=A' $ with the sequence in question being injective Finally, you construct a bijection $ g: S \longrightarrow S \setminus A $ with $ g(s)=s $ if $ s\in S \setminus A' $ and $ g(a_{m})=g(a_{m+n})$ $\forall m\in \mathbb{N^{*}} $

AlphaOmega
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