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I started real analysis and picked up a book called Understanding Analysis by Stephen Abbott. I understand why induction cannot be used prove the infinite case. From my understanding, induction proves that a statement is true for all natural numbers but not the infinite case because it only takes into consideration one case at a time (Is this correct?). Abbot uses the example $\bigcap^{n}_{i=1}A_i\ne \emptyset $ while $\bigcap^{\infty}_{i=1}A_i=\emptyset$ when $A_i$ is defined as a subset of $\mathbb{N}$ to show why induction would not work. Would strong induction suffice as a proof instead? What led me to this was how Abbott proved that for a set $A_n$=$\mathbb{N}\cap[n,\infty)$ and n $\in\mathbb{N}$, $\bigcap^{\infty}_{n=1}A_n=\emptyset$ using strong induction:

Suppose we had some natural number m which satisfies $m\in\bigcap^{\infty}_{n=1}A_n$. But since $m{\notin}A_{m+1}$, we have a contradiction so $\bigcap^{\infty}_{n=1}A_n=\emptyset$.

It seems as if strong induction uses all the previous cases to imply that the next case would be true so I was wondering if it would work for the infinite case. Apparently, transfinite induction extends to infinity which I learned from:

Why doesn't induction extend to infinity? (re: Fourier series)

and it seems very similar to strong induction such as the successor case which requires P($\alpha$) and sometimes P($\beta$) to be true for all $\beta\lt\alpha$ in order to prove P($\alpha$+1). This may be wrong though because I know nothing about ordinals. Here is the link: https://en.wikipedia.org/wiki/Transfinite_induction. Thank you for your time.

TrustinN
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    It turns out induction and strong induction (and also the well-ordering principle of $\Bbb{N}$) are equivalent, so if something can't be deduced using induction, it also can't be deduced using strong induction. – peek-a-boo Dec 31 '20 at 05:55
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    Induction is equivalent to strong induction. The proof that $\bigcap_{n=1}^\infty [n,\infty)=\varnothing$ does not use induction in either form, though. – pancini Dec 31 '20 at 05:55
  • @ Elliot G I thought it uses strong induction since it says that m is true for all sets from A_1,..., A_m and proves it is false for A_m+1. – TrustinN Dec 31 '20 at 05:58
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    For transfinite induction to work, we need to consider the case of limit ordinals too, and strong induction doesn't do that, so no, strong induction cannot reach infinite ordinals. As others have pointed out, though, something like $\bigcap_{n=1}^\infty A_n$ doesn't require an infinite case (even arbitrary intersections work without inducting through all the sets) – Brian Moehring Dec 31 '20 at 05:59

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The proof that $\bigcap_{n\ge 1}A_n=\varnothing$ when $A_n=\Bbb N\cap[n,\to)$ doesn’t use induction at all. It’s clear that the intersection is a subset of $\Bbb N$, since each of the sets $A_n$ is, so to show that the intersection is empty, we need only show that it contains no element of $\Bbb N$. And that follows immediately from the definition of intersection: for any $k\in\Bbb N$, $k\notin A_{k+1}$, so $k\notin\bigcap_{n\ge 1}A_n$.

By the way, strong induction is equivalent to ordinary induction: anything that can be proved using one can be proved using the other, though converting a proof by strong induction to a proof by ordinary induction typically makes it a little more complicated.

Brian M. Scott
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