[shkkmo]

> This claim is correct for all k > 1, but it is not correct for k=1. The problem is that we only show the truth of the proposition for N=1.

This is exactly what the person you are responding to was saying. They were talking about the transitive part of the inductive step that the article handwaves to with this line:

>> In particular, h_1 is brown. But when we removed h_1, we got that all the remaining horses had the same color as h_2. So h_2 must also be brown.

the "remaining horses" is the A ⋂ B set that was mentioned and if that set is empty then know it is the same color as set A is meaningless. Thus given f(A ⋂ B) = f(A) and f(A ⋂ B) = f(B), you only know that f(A ⋃ B) is constant if A ⋂ B is non-empty.

Given that in this case A and B are formed by removing different elements from a set of size n+1, the proof only works if n+1>=3 so that A ⋂ B can have at least one member.

[thaumasiotes]

> Thus given f(A ⋂ B) = f(A) and f(A ⋂ B) = f(B), you only know that f(A ⋃ B) is constant if A ⋂ B is non-empty.

But this is wrong. Given f(A ⋂ B) = f(A) and f(A ⋂ B) = f(B), you do not know that f is constant, regardless of whether A ⋂ B is or isn't a non-empty set. poetically described a completely different and obviously false claim instead of the actual claim made by the false proof.

[shkkmo]

You must not understand the notation. f(A ⋂ B) = f(A) is saying that every result of f(A ⋂ B) is equal to every result of f(A), this is trivially true if A ⋂ B is the empty set and thus tells you nothing. However if A ⋂ B is not empty, it follows from knowing f(A) is constant.

Poetically is correctly identified the exact step when the n>1 assumption is introduced. The argument is condensed but accurate.

> if f(A) = f(B) = f(A ⋂ B) then f is equal to some constant c and f(A ⋃ B) = c. This argument is valid only if A ⋂ B ≠ ∅.

It is very clearly the assumption that "all the remaining horses" is not empty that introduces the n>1 condition. It is the non emptiness of that set that allows you to say that since f(A) is constant and f(B) is constant (and we already know that because |A|=n and |B|=n), then f(A ⋃ B) is also constant.

[thaumasiotes]

You've gotten confused somewhere.

> However if A ⋂ B is not empty, it follows from knowing f(A) is constant.

That f(A) is constant is supposedly a conclusion, not a premise. It is not a valid conclusion to draw from the stated premises.

> f(A ⋂ B) = f(A) is saying that every result of f(A ⋂ B) is equal to every result of f(A)

As I just responded above, interpreting the claim this way doesn't get it to make any more sense. When f(A ⋂ B) = f(A), there is no basis from which to conclude that f is constant. You have no information about whether f is or isn't constant.

This is not the claim made in the false proof, nor does it have anything to do with the claim made in the false proof. It is a creation of poetically's own mind, and it makes no sense.

[shkkmo]

> You have no information about whether f is or isn't constant

You seem to have forgotten what we are talking about. A ⋃ B is an arbitrary set of horses of size n+1, A and B are subsets of A ⋃ B, each with a single different element removed and thus of size n. We already know that f() is constant over A and over B because they are of size n and we have assumed that all sets of horses of size n have the same color (which I already pointed out in my last comment.) We also know that f() is constant over A ⋂ B becasue it is a subset of sets we already know f() is constant over. The only thing that needs to be proven is that f() is constant over the set A ⋃ B. That proof is impossible if A ⋂ B is empty. We can only assert A ⋂ B is not empty if we assume that A ⋃ B must have a size of at least 3. From assuming n+1>=3 we are able to correctly conclude that n>1 implies the inductive step is valid.