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Let $P_n, n\in\mathbb N$ a property. We suppose that $P_{n_0}$ is true for a certain $n_0\in \mathbb N$ and that $P_n$ true implies $P_{n+1}$ true for all $n\geq n_0$. Prove that $P_n$ is true for all $n\geq n_0$.

Attempts

Suppose it's no true, and Let $m>n_0$ s.t. $P_m$ is false. Let $A=\{n\mid P_n\ false\}\subset \mathbb N$. Since $\mathbb N$ is well ordered and that $A\neq\emptyset$, there is a minimal element $k=\min A$. In particular, $P_{k-1}$ is true and $P_{k}$ is wrong which is a contradiction with the fact that $P_{n}\implies P_{n+1}$ for all $n$.

Therefore $P_n$ true for all $n$.

Question : Is it true ? My problem is that I don't understand where I used the fact that $P_{n_0}$ true.

user380364
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1 Answers1

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If there is no $n$ such that $P_n$ true, then $0=k=\min A$ and thus $P_{k-1}$ wouldn't have any sense.

Surb
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  • We're given that $P_{n_0}$ is true. – Kevin Sep 16 '17 at 18:46
  • @Kevin: You should read the question ;-) – Surb Sep 17 '17 at 05:32
  • It's ambiguous, but seeing as the OP wrote "we suppose" before the imperative "Prove that...", I read it as a given rather than an assumption. – Kevin Sep 17 '17 at 05:34
  • @Kevin : it's not ambiguous at all, and the question is precisely : Where did I used the fact that $P_{n_0}$ is true. An I answered to this question. – Surb Sep 17 '17 at 06:15
  • Ah, I see. You should answer the question (with a "yes" or "no"). – Kevin Sep 17 '17 at 06:24