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?
You should not be using $|A|$ anywhere since $|A|=\infty$. However your algorithm for constructing the sequence is fine. After describing it (step j) you can claim the sequence is monotonic increasing by construction. Then you use the infinite size of $A$ to ensure that there always exists a $n_{j+1} > n_j$ because the set $A\setminus \{ n_1 , n_2, ...,n_{j} \}$ is also infinite.