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I have no clue how to do this starting from the 3rd statement:

N3: 1 is not a successor of any $n \in \mathbb{N_F}$. I tried by contradiction, so then $0 \in N_F$ because $F$ is a field, where is the contradiction? $0<1$ sure but how does that contradict anything?

Another problem I have is N5, so let $S \subseteq N_F$. Assume $S$ satisfies $1 \in S$ and if $n \in S$ then $n+1 \in S$. Then I guess I'm supposed to prove $N_F \subseteq S$. To do this I'd pick an arbitrary $n \in N_F$. Then what if $n = 1.5$, there is nothing saying that $n$ cannot be 1+1/2, where $1/2 \in N_F$. What do I do? Everything seems easy once I can show that $N_F$ is no more than all $\mathbf{n}$, but I have no idea how to do so.

Also when I apply induction on $N_F$, I could show that $\forall \mathbf{n}, \mathbf{n} \in N_F$. Though this doesn't prove that there are no other elements in $N_F$ except for $\mathbf{n}$.

nabu1227
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  • The issue is the phrase "define recursively" - what exactly does that mean? Certainly $\mathbb{N}$ isn't the only set containing $1$ and closed under $x\mapsto x+1$, so there's got to be some other piece of the definition - see e.g. here. – Noah Schweber Jan 24 '21 at 07:37
  • I see this problem. However, I cannot find a way to show that if I choose an arbitrary element in the field that is not $\mathbf{n}$, then it does not belong in $N_F$. – nabu1227 Jan 24 '21 at 07:39

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