I suppose we cannot say "Let natural numbers be $1, 1+1, 1+1+1, \dots$", but I fail to put into words why we can't. What's the reason to define natural numbers as an intersection of inductive sets? If we could define it as I wrote above, we wouldn't be troubled with questions like "Is there a natural number between $1$ and $2$?"
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4Well, what do you mean by "..."? – Qiaochu Yuan Sep 21 '22 at 06:57
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@QiaochuYuan I mean we proceed by adding $1$ to the previous number, in question the next will be $1+1+1+1$ and so on – Alexander Vronsky Sep 21 '22 at 07:03
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3@AlexanderVronsky What you described is precisely an inductive definition of natural numbers... – 5xum Sep 21 '22 at 07:07
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@5xum True. Then why we trouble ourselves with an intersection of all inductive sets instead of specifying the one that matters? – Alexander Vronsky Sep 21 '22 at 07:12
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@AlexanderVronsky Well, provided that you have an axiom that says that there is an inductive set (i.e. a set $s$ such that $1\in s\land \forall z, (z\notin s\lor z+1\in s )$ ), you are specifying the one that matters: $x\in \Bbb N\Leftrightarrow (\forall y, (\neg\text{Inductive}(y)\lor x\in y))$. If you haven't got that axiom, then the probem is that (provided giving a set-theoretical meaning to $1$ and $y+1$) you cannot prove that there is a set $X$ such that $1\in X\land\forall y,(y\notin X\lor y+1\in X)$. – Sassatelli Giulio Sep 21 '22 at 07:26
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1@AlexanderVronsky We add finite sums of the form $\sum 1$ to our candidate set. And doing that any finite number of times is fine, we get a set. You might protest with "well, just add such elements indefinitely". Fair enough, but how do you know that this process yields a set? An infinite set, at that. You have to assume it does, there is no other way. – AlvinL Sep 21 '22 at 08:00
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@SassatelliGiulio okay, I guess I understand. Thank you – Alexander Vronsky Sep 21 '22 at 08:05
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@AlvinL Also true, thank you very much! – Alexander Vronsky Sep 21 '22 at 08:24