**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( Asetminus { n_1 }) $

step 3: $ n_3 = min(Asetminus { n_1 , n_2 }) $

$ vdots$

step j: $ n_j = min(Asetminus { 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(Asetminus { 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?