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I was recently asked to help someone with the following question on their first year analysis course.

Recall that $\mathbb{N}$ is the set of all positive integers. Use the principle of induction to show that $n \ge 1$ for all $ n \in \mathbb{N}.$ [ Hint: Let $ S = \lbrace n \in \mathbb{N} | n \ge 1 \rbrace.$ ]

I am concerned that the first sentence of the question assumes that we already know exactly what the positive integers are, which renders the rest of the question redundant.

I would prefer to see the question written:

Recall that $\mathbb{N}$ is the set of $ x \in \mathbb{R}$ such that $x$ is a member of every inductive set. [ $P$ is an inductive set if (a) $ 1 \in P $ and (b) $ x \in P \Rightarrow x+1 \in P.$ ] Then we can answer as follows:

With $S$ as in the hint, by definition $ S \subseteq \mathbb{N}.$

$1 \in S$ since $1 \in \mathbb{N},$ as it is an inductive set. For $ n \in S, $ $n+1 > n \ge 1,$ therefore $n+1 \in S.$ Hence $S$ is an inductive set, and so $ \mathbb{N} \subseteq S.$ Therefore $ S=\mathbb{N}.$ Therefore $1$ is the least element of $\mathbb{N}.$

Am I being way too picky, or is the question flawed as it stands?

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    "Principle." Whether we know what the natural numbers "are" doesn't necessarily mean we know that they have this property. It all depends on what definitions they started out with and, moreover, it doesn't really matter. (Also, R? What?!)2010-10-29
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    @Qiaochu That's exactly my point; the definition that we start with. Without that, things look kind of circular to me.2010-10-29
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    If the lecturer previously defined $\mathbf{N}$ as the intersection of all inductive subsets of $\mathbf{R}$ then this is perfectly fine. (If I recall correctly, this is what Munkres does in his celebrated textbook on topology.)2010-10-29
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    @Robin Chapman Yes, this is in line with what I was thinking but I was slightly sceptical over how $\mathbb{N}$ was defined (I don't have this information) given that it came up just a few weeks into a first year course. I guess I shouldn't jump to conclusions.2010-10-29

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