Proof By Least Element Axiom

Question

Suppose I have some finite summation $P(n)$ on $[0, n]$, where $n\in \mathbb{N}$. I want to prove a particular formula describing it. In my structures course I'm taught to assume that with a set $S$ for all $k\in \mathbb{N}$, $P(k)$ does not equal said formula.

But suppose I find a base case where the formula is satisfied. Suppose $P(0)$ follows whatever proposition I'm given. At this point I say that $k=0 \notin S$, therefore there must be a "least element $l$" not within set $S$. If $l\ne0$ then there should be an $l-1$ not within $S$

After some algebra is done, say it turns out that the summation with upper limit $l-1$ turns out to agree with the proposition describing $P$, therefore there is no least element within $S$, proving by contradiction that the set $S$ is false for all $k$, therefore $P(n)$ is true for all $n\in\mathbb{N}$

Now,

• why does it matter if the base case at $k=0$ agrees with the proposition? I understand if it turns out it's false, that means the entire statement is false because of the universal quantifier. But what if it turns out it's a coincidence the base case is true?
• I still don't understand exactly what is meant by a "least element". Does this mean smallest element or at least 1 element? If it means at least 1 element, then would I be able to make l any element in the middle of our domain?
• If our domain is the set of naturals, then why does $0$ have any bearing on our argument?
• What does $l-1$ signify? Why is it not $l+1$ like I normally see in induction?

Answer 1

Let me answer the two simple questions first. Least element means smallest element. And for many mathematicians, including me, the natural numbers include $0$: $\Bbb N$ is the set of non-negative integers. That is evidently the way the term is being used in your course or text.

The other two questions are, I think, best answered by examining the structure of this kind of proof.

Suppose that $P(n)$ is a statement about natural numbers $n$, and we want to prove that $P(n)$ is true for all natural numbers, i.e., for all integers $n\ge 0$. One way to do so is to show that $P(0)$ is true and then suppose that there is some $n\in\Bbb N$ such that $P(n)$ is not true. In that case we can let

$$S=\{n\in\Bbb N:P(n)\text{ is false}\}$$

and know that $S\ne\varnothing$. The Least Element axiom then tells us that $S$ has a least, i.e., smallest, element. That is, there is an $\ell\in S$ such that $\ell\le n$ for each $n\in S$.

We proved that $P(0)$ is true, so we know that $\ell$ cannot be $0$: we must have $\ell>0$, and therefore $\ell-1\in\Bbb N$. But $\ell$ is the smallest number in $S$, so $\ell-1\notin S$, and therefore $P(\ell-1)$ must be true: if it were false, $\ell-1$ would have to be in $S$.

We can then complete the proof by showing that if $P(\ell-1)$ is true, then $P(\ell)$ must also be true, because that will give us a contradiction: by definition $\ell\in S$, so $P(\ell)$ isn’t true. This contradiction was caused by the assumption that $S\ne\varnothing$, i.e., that $P(n)$ is false for some $n\in\Bbb N$.

You can regard this as a justification of mathematical induction: it’s a proof that if every non-empty set of natural numbers has a least element, then proof by induction is legitimate. Conversely, one can prove that if proof by induction is legitimate, then every non-empty set of natural numbers has a least element. They’re two ways of looking at the same thing, and they require you to do exactly the same things in carrying out a proof: prove that $P(0)$ is true, and prove that for any $n\in\Bbb N$, $P(n)$ implies $P(n+1)$.

If you think in terms of the usual elementary presentation of induction, you probably think of $P(0)$ implying $P(1)$, which then implies $P(2)$, and so on, until ‘all of the dominoes have fallen’.

If you think in terms of the Least Element axiom, on the other hand, you say to yourself that if the theorem were false, there would be a smallest counterexample — $\ell$ in the discussion above. Then you’d argue that this is impossible, because $\ell$ can’t be $0$ — you’ve checked that $P(0)$ is true — but then $\ell-1$ is a natural number for which the theorem is true, and $P(\ell-1)$ implies $P(\ell)$, so $P(\ell) is true. OOPS!! That’s a contradiction, so there must not be any counterexamples after all, and the theorem must be true.