# A challenge to constructivists

Constructive mathematicians do not accept a proof of existence unless it provides a recipe for how to construct the thing whose existence is being asserted. Constructive mathematics is quite interesting, but it also appears to have some big problems. Here’s a challenge for constructivists:

Suppose that I hand you some complicated function f from a set A to another set B. I ask you: “Can every element in B be reached by applying the function to an element in A?” In other words, is f surjective?

Now, it so happens that the cardinality of B is greater than the cardinality of A. That’s sufficient to tell us that f can’t be surjective, as however it maps elements there will always be some left over. So we know that the answer is “no, we can’t reach every element in B.” But we proved this without explicitly constructing the particular element in B that can’t be reached! So a constructivist will be left unsatisfied.

The trick is that I’ve made this function extremely complicated, so that there’s no clever way for them to point to exactly which element is missing. Would they say that even though |B| is strictly larger than |A|, it could still be somehow that every element in B is in the image of f? Imagine asking them to bet on this proposition. I don’t think any sane person would put any money on the proposition that f is onto.

And as a final kicker, our sets don’t even have to be infinite! Let |A| = 20 and |B| = 21. I describe a function from A to B, such that actually computing the “missing element” involves having to calculate the 21st Busy Beaver number or something. And the constructivist gets busy searching for the particular element in B that doesn’t get mapped to, instead of just saying “well of course we can’t map 20 elements to 21 elements!”

Even simpler, let F map {1} to {1,2} as follows: F(1) = 1 if the last digit of the 20th busy beaver number is 0, 1, 2, 3, or 4, and F(1) = 2 otherwise. Now to prove constructively that there is an element in {1, 2} that isn’t in the image of F requires knowing the last digit of the 20th busy beaver number, which humans will most likely never be able to calculate (we’re stuck on the fifth one now). So a constructivist will be remain uncertain on the question of if F is surjective.

But a sane person would just say “look, of course F isn’t surjective; it maps one object to two objects. You can’t do this without leaving something out! It doesn’t matter if we don’t know which element is left out, it has to be one of them!”

And if humanity is about to meet an alien civilization with immense computational power that knows all the digits of the 20th busy beaver number, the standard mathematician could bet their entire life savings on F not being surjective at any odds whatsoever, and the constructive mathematician would bet in favor of F being surjective at some odds. And of course, the constructivist would be wrong and lose money! So this also means that you have a way to make money off of any constructivist mathematicians you encounter, so long as we’re about to make contact with advanced aliens.

## 2 thoughts on “A challenge to constructivists”

1. Anonymous says:

did you mean, in the third paragraph, “Now, it so happens that the cardinality of B is greater than the cardinality of A.”?