Problem: If $ A \subseteq \mathbb{N} $ is an infinite set then there exists a strictly monotonic increasing sequence $ (n_k)_{k=1}^\infty $ s.t. $ A = \{ n_k | k \in \mathbb{N} \} $
My attempt of proof:
Perform the following algorithm,
step 1: Define $ n_1 = minA $
step 2: $ n_2 = min( A\setminus \{ n_1 \}) $
step 3: $ n_3 = min(A\setminus \{ n_1 , n_2 \}) $
$ \vdots$
step j: $ n_j = min(A\setminus \{ n_1 , n_2, ...,n_{j-1} \}) $
We'll perform $ |A| $ steps and define each time $ n_i $ for all $ 1 \leq i \leq |A| $.
Noticing that $ n_1 < n_2 < ... < n_{ |A| } $ ( we have a strictly monotonic sequence ) and that $ n_i \in A $ for all $ 1 \leq i \leq |A| $ so we're finished.
More compact attempt of proof:
Define $ n_1 = minA $ . For all $ n_i \in A $ s.t. $ 2 \leq i \leq |A| $, we'll define $ n_i = min(A\setminus \{ n_1 , n_2, ...,n_{i-1} \}) $
Noticing that $ n_1 < n_2 < ... < n_{ |A| } $ and that $ n_i \in A $ for all $ 1 \leq i \leq |A| $ so we're finished.
I don't know if I'm correct since $ A $ is an infinite set and I haven't seen proof by algorithms in which the iteration is continuing infinitely ( there are $ |A| $ iterations in the algorithm above ), neither I have seen the usage of indexing on an infinite set ( for example, I have wrote: " for all $ 1 \leq i \leq |A| $ , but $ |A| $ is not a finite number since $ A $ is infinite set " ). Are my proofs correct? if not, what is the problem with them?