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Here there is a proof
but I think it is incomplete(missing $1\in\Bbb{N}$)


Note: A⊆R is inductive if and only if 1∈A and ∀x∈A⇒x+1∈A.
By definition, $\Bbb{N}$ is the intersection of all inductive sets

I tried to do a proof here:

$\Bbb{N} = \bigcap A_i\in I$ : $1 \in A_i \land \forall x\in A_i \Rightarrow x+1 \in A_i$

This is the same as:

$\Bbb{N} = \{x: x\in A_1 \land x\in A_2 \land x \in A_3 \land ...\}$

As $1 \in A_i\in I \Rightarrow 1\in \bigcap A_i\in I $ ie $1\in \Bbb{N}$

the rest is same

Is my proof correct?

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  • $\begingroup$ 1 is in N by definition of intersection. 1 is in all A_i so 1 is in the intersection. $\endgroup$
    – fleablood
    Jan 31, 2016 at 3:47
  • $\begingroup$ The set of inductive sets is not countable. So ... no ... I don't think your proof works. $\endgroup$
    – fleablood
    Jan 31, 2016 at 3:49
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    $\begingroup$ Leave off the indexes. Simple say N = $\cap A; A \in I$ and $\forall A \in I; 1 \in A$ and therefore $1 \in N =\cap A$. Then your proof is good. $\endgroup$
    – fleablood
    Jan 31, 2016 at 3:56

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