Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am currently studying analysis and I came across the following exercise.

Proposotion 2.2.14 Let $m_0$ be a natural number and let $P(m)$ be a property pertaining to an arbitrary natural number $m$. Suppose that for each $m\geq m_0$, we have the following implication: if $P(m')$ is true for all natural numbers $m_0\leq m'< m$, then $P(m)$ is also true. (In particular, this means that $P(m_0)$ is true, since in this case the hypothesis is vacuous.) Then we can conclude that $P(m)$ is true for all natural numbers $m\geq m_0$.

Prove Proposition 2.2.14. (Hint: define $Q(n)$ to be the property that $P(m)$ is true for all $m_0\leq m < n$; note that $Q(n)$ is vacuously true when $n<m_0$.)

I have difficulty understanding how I should use the hint and in general what the framework of this proof would look like (probably an inductive proof; but on what variable do we induct, what will be the induction hypothesis and how would I go about proving the inductive step etc.?). Could anyone please provide me with some hints to help me get started?

share|cite|improve this question
Should the last sentence have $m\ge m_o$? – Stefan Hamcke Jul 30 '13 at 13:37
@StefanH. Yes, you're right. Silly typo. – dreamer Jul 30 '13 at 13:40
Exact what definition do you have for natural numbers? Or can you rely on the (weak) induction principle? – skyking Sep 7 '15 at 7:53
up vote 2 down vote accepted

Let $B$ be the subset of $N=\{m_0,m_0+1,...\}$ such that $P(m)\iff m\in B$. This $B$ is not empty since for all $m_0\le m'<m_0$ the property $P$ is satisfied, thus also $P(m_0)$. We want to show that $B=N$. So assume that $A:=N\setminus B\ne\emptyset$. Then there is an $m\in A$ such that each $m_0\le n<m$ is in $B$, in other words $m$ is the minimum of $A$. But if $n<m$ implies $P(n)$, then by hypothesis $P(m)$ and so $m\in B$, a contradiction. Hence $B=N$.

Remark: This works for all sets $N$ where each non-empty subset has a minimal element with respect to a relation $R$. These sets are called well-founded.

If you want to use the hint, show that $Q(n)$ implies $Q(n+1)$ and that $Q(m_0)$: Since $Q(m')$ is true for all $m_0\le m'< m_0$, it is also true for $m_0$. Assume $n$ is a natural number $\ge m_0$ such that $Q(n)$. This means that $P(m)\ \forall m_o\le m<n$. By hypothesis this implies $P(n)$, thus $P(m)\ \forall m_0\le m<n+1$, so again by definition of $Q$ we have $Q(n+1)$. Now apply the induction principle.

So we can proof the strong induction principle via the induction principle. However, the normal induction principle itself requires a proof, it that is the proof I wrote in the first paragraph. As mentioned it works for all well-founded sets ($\mathbb N$ is such a set.)

share|cite|improve this answer
Thanks a lot for your help! One thing that isn't clear to me yet though is how you incorporated the hint that was given in the question in your answer. Or is the hint just not necessary at all? – dreamer Jul 30 '13 at 13:53
I proofed the strong induction principle using only the fact that each non-empty subset has a minimal element. The hint isn't useless, however, since it can be used to proof the strong induction principle from the (weak) induction principle. I'll edit that into my answer... – Stefan Hamcke Jul 30 '13 at 14:02
@rbm: No I used the original Induction Principle ($(Q(n)\to Q(n+1))\implies Q(m) \forall m\in\Bbb N$) and the hypothesis $P(n)\forall m_0\le n<m \to P(m)$. – Stefan Hamcke Jul 30 '13 at 14:17
Your edit clarified it :). Thanks again :)! – dreamer Jul 30 '13 at 14:18
Sorry to pull this out of the archives, but I had a query regarding the note in the hint. Can you please try to clarify what did Prof. Tao intend to say there? – Aseem Dua May 23 '14 at 7:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.