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I have difficulty understanding part of the comment mentioned after the following theorem:

Let $A \subset \mathbb {N}$ be such that, for every natural number $n$, if every $m < n$ satisfies that $m \in A$, also $n \in A$. So $A = \mathbb {N}$.

The part that I have difficulty understanding is highlighted in bold

Note that with this formulation it is not necessary to require a proof apart from $P0$, because it is trivially true that "every natural number less than that $0$ satisfies the property $P$”, and we are assuming that under this trivial case from the induction hypothesis we can prove $P0$. This does not mean that, in practice, assuming that every natural number less than n satisfies P, it may be convenient to treat separately the cases $n = 0$ (where the hypothesis of induction does not contribute anything) and $n \not= 0$.

I appreciate in advance if someone can explain to me what that part of the text refers to.

Oscar F Padilla
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  • The quoted text (both the statement of the theorem and the comment) is confusingly written. "This does not mean that it may be convenient" ought to be "this does not mean it is never convenient", i.e. it may indeed be convenient. Anyway, are you asking why the induction hypothesis does not contribute anything when $n=0$? That's because the induction hypothesis is vacuously true in that case since there are no numbers less than $0$. – Karl Jul 17 '23 at 03:44
  • Also, I wouldn't dwell on this too much - just look at some examples. – Karl Jul 17 '23 at 03:45
  • Thanks for answering, my question is "in which cases it is convenient to separate the cases when $n=0$" (according to the text it is when "the induction hypothesis does not contribute anything", but this is something that I do not understand what it means either) – Oscar F Padilla Jul 17 '23 at 03:51
  • In fact, $A=\emptyset$ satisfies the condition. – ultralegend5385 Jul 17 '23 at 03:54
  • It just means that when $n=0,A=\emptyset$, the inductive hypothesis is trivial. I highly recommend just looking at some examples. – Eric Jul 17 '23 at 03:55
  • Original question didn't have $\mathbb{N}$--which makes it fairly obviously an incorrect assertion. Otherwise it appears to suggest that you can choose any natural number and all previous natural numbers must be part of the set. You can show this by assuming it's true for $m \leq n$ and thus true for $m < n$: take the base case of 1 and 2 to start the chain (inductive reasoning). – Jared Jul 17 '23 at 04:04
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    See discussions here and here. – Arturo Magidin Jul 17 '23 at 05:07

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