I am currently studying analysis and I came across the following exercise.
Proposotion 2.2.14
Let m0 be a natural number and let P(m) be a property pertaining to an arbitrary natural number m. Suppose that for each m≥m0, we have the following implication: if P(m′) is true for all natural numbers m0≤m′<m, then P(m) is also true. (In particular, this means that P(m0) is true, since in this case the hypothesis is vacuous.) Then we can conclude that P(m) is true for all natural numbers m≥m0.
Prove Proposition 2.2.14. (Hint: define Q(n) to be the property that P(m) is true for all m0≤m<n; note that Q(n) is vacuously true when $n
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?
Answer
Let B be the subset of N={m0,m0+1,...} such that P(m)⟺m∈B. This B is not empty since for all $m_0\le m'
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(m0): Since Q(m′) is true for all m0≤m′<m0, it is also true for m0. Assume n is a natural number ≥m0 such that Q(n). This means that $P(m)\ \forall m_o\le m
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 (N is such a set.)
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