The theorem that I am going to attempt to prove is that every infinite set has a countably infinite subset. Someone explain to me without a bald assertion what's wrong with this proof. I use strong induction. Let S be the infinite set.

First: let x_{1} be a member of S.

Second: Assume that *x _{k}*, for all k from 1 to

*n*, exist, are members of S, and are distinct. Let

*x*be a member of S - U

_{k+1}*{*

^{n}_{k=1}*x*} (meaning S without

_{k}*x*...

_{1}*x*). Then

_{n}*x*, for all

_{k}*k*from 1 to

*n*+1, exists, are members of S, and are distinct.

Conclusion: *x _{n}* exists for all positive integers

*n*. They are distinct members of S. Their union, U

_{k=1}

*{*

^{infinity}*x*} is a countable subset of S. QED.

_{k}Again, someone tell me what's wrong with this proof, without a bald assertion. The bald assertion I'm thinking of is that some version of the Axiom of Choice is needed to get a full infinite set of elements *x _{n}*. Proof Wiki gives just this warning. However, this method proves that

*x*exists for all positive integers

_{n}*n*, and only uses only a finite number of

*x*values to get each new

*x*value.

At least, they cite the ZF set theory for this assertion. I should learn and understand the theory.

Someone? I'm listening.

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