I think that what I wrote is implicitly agreeing with the GP that the induction argument is wrong not the base case. And I think that you attempted to refute it, but I couldn’t really understand how what you wrote would relate to either side. Do you mean that you agree that your description above was complicated, or that you agree that what I wrote is a fair description of the problem, or do you (now as opposed to before?) think that the problem was the induction step not the base case? Or are you talking about the statement that it is easy to miss the error by thinking of n as big rather than n=2? Maybe it doesn’t matter.
The argument in the article is the following: Given some function f, 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 doesn't apply to the base cases because anything is true for the empty set and any function on a one element set is trivially constant. So the minimal case where this can be applied in a non-trivial way is a set with 3 elements.
> The argument in the article is the following: Given some function f, if f(A) = f(B) = f(A ⋂ B) then f is equal to some constant c and f(A ⋃ B) = c.
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...
I've constructed this function so that, obviously, f(A) = f(B) = f(A ⋂ B) = 5, and A ⋂ B = [1,2] ≠ ∅, but f is not a constant function and f(A ⋃ B) is not 5. f is still a function.
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 )
where f is the function that tells you what color a horse is, and N is a free variable.
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 )
We can now say that the proof is claiming that whenever p(k) is true, p(k+1) is also true, that this claim is correct for all k > 1, and that p(1) has been established. But p(2) has not.
> 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.
> 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.
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.
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.
Under your interpretation, poetically's claim still does not reflect the claim from the false proof, and it is still itself obviously wrong.
(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]
We can easily see that f(A) = f(B) = f(A ⋂ B) = A. But it is not true that f takes a constant value over any of those domains. (It is true that f(A ⋃ B) = f(A), but this is required by the premise f(A) = f(B).)
In your example f(A) is not a singleton. I think that the image f(A) being a singleton {c} is what was meant by “f(A) is some constant c”.
It feels to me like you are taking advantage of imprecise language to demonstrate counter-examples to statements that no one was trying to prove.
Do you actually think that 'poetically (or I) are trying to use obviously false properties of functions (specifically the properties you keep providing counter-examples to) or are just being nit-picky about language?
Usually I find much mathematical writing can be hard to read because much of the context or scope of variables, assumptions, or definitions is so implicit. But when something seems obviously wrong it is an exercise to figure out what constraint I’ve missed that makes it true.
When one first learns analysis, the statement of so many of the theorems begins “given a continuous function f” and this continues in formal writing but when people communicate more casually they usually just say function and everyone knows/figures that a continuous function is meant instead of pointing out an obvious counter-example.
In the context of this thread, it is obvious to me that the thing being talked about is when the image of a function restricted to some subset of the domain is a singleton. The reasons it is obvious to me are:
- The f(A) := {f(a) | a in A} notation is standard and implied by the choice of letters
- It is meaningless to say that the value of a function is constant—you always get the same result, so for the word to mean something, I think it is reasonable to take it as implying the restriction is a constant function, or in other words that f(A) = {c}, a singleton.
- It seemed obvious to me from the context of the article
- It seems a reasonable interpretation that changes the statements from nonsense into a reasonable description
At first I thought you just weren’t familiar with the notation, but now it seems you knew it so was it just not obvious to you that that is what was meant? Or do you actually think people were trying to prove or rely on the statements you disproved? Or that people should write more precisely and deserve to be punished with pedantry when they assume some mathematical maturity instead? I’m genuinely curious what you think is going on here because I’m struggling to fit these comments to a model of how hn commenters think and act.
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.
I think that what I wrote is implicitly agreeing with the GP that the induction argument is wrong not the base case. And I think that you attempted to refute it, but I couldn’t really understand how what you wrote would relate to either side. Do you mean that you agree that your description above was complicated, or that you agree that what I wrote is a fair description of the problem, or do you (now as opposed to before?) think that the problem was the induction step not the base case? Or are you talking about the statement that it is easy to miss the error by thinking of n as big rather than n=2? Maybe it doesn’t matter.