Questions on "All Horse are the Same Color" Proof by Complete Induction

Solution 1:

It’s clear from the question and from your discussion with @DonAntonio that you don’t actually understand the induction step of the argument. You seem to think that it involves comparing some previously chosen set of $n$ horses with some new set of $n+1$ horses, but it does not. Let me see if I can explain more clearly just how it does work.

We assume as our induction hypothesis that if $H$ is any set whatsoever of $n$ horses, then the horses in $H$ are all the same color. Now let $H=\{h_1,h_2,\dots,h_{n+1}\}$ be any set of $n+1$ horses; our goal is to show that all of the horses in $H$ are the same color. To do this, we look at the subsets

$$H_0=\{h_1,\dots,h_n\}\quad\text{and}\quad H_1=\{h_2,\dots,h_{n+1}\}\;.$$

Each of these subsets contains $n$ horses, so by hypothesis all of the horses in $H_0$ are the same color, and all of the horses in $H_1$ are the same color.

Note: The two $n$-horse sets that we’re looking at here were not given to us ahead of time: they are just two different $n$-horse subsets of $H$. And in fact any two different $n$-horse subsets of $H$ will work equally well for the argument.

Now if $n\ge 2$, the horse that I’ve called $h_2$ is in both $H_0$ and $H_1$. (In fact every horse except $h_1$ and $h_{n+1}$ is in both $H_0$ and $H_1$.) Since $h_2$ is in $H_0$, and we already know from the induction hypothesis that all of the horses in $H_0$ are the same color, it follows that every horse in $H_0$ is the same color as $h_2$. But $h_2$ is in $H_1$ as well, so by exactly the same reasoning we conclude that every horse in $H_1$ is the same color as $h_2$. But every horse in $H$ is in $H_0$ or $H_1$ (or both), so every horse in $H$ is the same color as $h_2$, and therefore the horses in $H$ are all the same color. This completes the induction step.

There is absolutely nothing wrong with that argument — provided that $n\ge 2$. In particular, it does not involve starting with some particular set of $n$ horses known to be the same color and trying to use that set to show that the horses in some arbitrary set of $n+1$ horses are all the same color. The argument starts with a set of $n+1$ horses and proceeds by looking at two $n$-horse subsets of the given set. It’s important here to realize that the induction hypothesis does not say that there is some particular set of $n$ horses that are all the same color: it says that in any set of of $n$ horses, all of the horses are the same color.

The argument fails only when $n=1$. In that case it fails because the sets $H_0=\{h_1\}$ and $H_1=\{h_2\}$ no longer overlap: there is no horse in both sets. Thus, while it’s true that all of the horses in $H_0$ are the same color and that all of the horses in $H_1$ are the same color, we can no longer infer that those two colors must be the same. But if we could somehow show that in every set of two horses, both horses were the same color, the induction argument as given would work just fine: we’d always be considering some $n\ge 2$, so the sets $H_0$ and $H_1$ would overlap.

Solution 2:

I'm not sure I understand fully your doubt, but the actual problem with that, and many other examples of bogus inductive proofs out there, is that when you take the two sets

$$\{h_1,...,h_n\}\;,\;\;\{h_2,...,h_n,h_{n+1}\}$$

each of size $\,n\,$ and thus, by the Ind. Hypothesis, formed of horses of the same color, the leap to deduce all the horses altogether have the same color requires that the intersection of both sets above is not empty, something you can't prove (because it is false!) if $\,n=2\,$

Thus, if we had the claim "every two horse whatsoever are of the same color", then we could prove "all the horses are of the same color"

Solution 3:

There is a more basic problem here : induction is irrelevant to the proof.

$P(2)$ is, by itself, a formalization of the statement that all horses are the same color. There is no reason to prove $P(2)$ to prove $P(3)$ to prove ... $P(n)$ and conclude from this chain of implications that all horses are the same color; one would try to prove $P(2)$ and in case of success, summarize that as "all horses have the same color".

If proving $P(n)$ were of interest for some reason, it can be derived in one step from $P(2)$, with no need for an induction.

The size of the set of horses is usually ignored in presentations of this argument and ones like it. But if there are $n$ horses in the universe, $P(k)$ is an empty statement for $k>n$, and is true regardless of the truth value $P(n)$. This shows that $P(n)$ is logically stronger than $P(n+1)$, and that the inductive step from $P(n) $ to $P(n+1)$ is vacuous for large $n$.