This is not the argument.
> This argument is valid only if A ⋂ B ≠ ∅.
No, it isn't; it isn't even valid then. Consider a function over sets of real numbers:
f([0,2]) = 5
f([1,3]) = 5
f([1,2]) = 5
f([0,3]) = 6
plus other values...
We can state the proposition "in a group of N or fewer horses, all of the horses are the same color" like so:
∀G∃c∀x( (|G| ≤ N ∧ x ∈ G) → f(x) = c )
The proof claims that if this proposition is true for the value N = k, then it is also true for the value N = k+1. 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.
Abstracting a little further, we can view the proposition above as the potential output of a function of N:
p(0) = ∀G∃c∀x( (|G| ≤ 0 ∧ x ∈ G) → f(x) = c )
p(1) = ∀G∃c∀x( (|G| ≤ 1 ∧ x ∈ G) → f(x) = c )
p(2) = ∀G∃c∀x( (|G| ≤ 2 ∧ x ∈ G) → f(x) = c )
p(3) = ∀G∃c∀x( (|G| ≤ 3 ∧ x ∈ G) → f(x) = c )
p(4) = ∀G∃c∀x( (|G| ≤ 4 ∧ x ∈ G) → f(x) = c )
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.
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.
I think you just misunderstood the notation. f is a function from (say) X -> Y. But the parent talks about it as if it were a function from the power set P(X). That is, for A a subset of X, when the commenter above says that f(A) is a constant, they meant that f(A) = {c} for some c in X. This is a bit of loose usage of the standard notation f(A) := {f(a) : a in A} is commonly understood.
Obviously an arbitrary function f : P(X) -> … needn’t satisfy the properties but no one was suggesting that.
(I'll assume that when you say "f(A) = {c} for some c in X", you mean "f(A) = {c} for some c in Y", since the values of f come from Y.)
So consider these definitions:
f(x) = x²
A = [0,1]
B = [-1, 1]
This is incorrect. You attribute an argument to the article that it does not make.
> This argument is valid only if A ⋂ B ≠ ∅
This is vacuously true in that the argument is not valid and its validity therefore implies all propositions including that A ⋂ B ≠ ∅. Although I tend to suspect that that isn't what you meant to say. It is equally true that "This argument is valid only if A ⋂ B = ∅". It is not true that if a function f takes a constant value, you can conclude that A ⋂ B ≠ ∅ for arbitrary A and B.