(4/6) Now here is a fantasy of how to construct an explicit counterexample.

We build two equivalence relations 𝑅 and 𝑆 on ℕ in such a way that the quotients ℕ/𝑅 and ℕ/𝑆 are both streamless but their cartesian product is not. To ensure the latter, we take care that 𝑅 and 𝑆 will be *orthogonal*: 𝑥𝑅𝑦 ∧ 𝑥𝑆𝑦 ⇒ 𝑥 = 𝑦. That way the diagonal function 𝑛 ↦ ([𝑛], [𝑛]) will expose that ℕ/𝑅 × ℕ/𝑆 is not streamless.

We start by setting each of 𝑅 and 𝑆 to be the (reflexive-symmetric-transitive closure) of the empty set.

And then we leave 𝑅 and 𝑆 that way.

Until somebody challenges us, by presenting a function 𝑓 : ℕ → ℕ and expecting us to come up with numbers 𝑖 < 𝑗 such that 𝑓(𝑖) 𝑅 𝑓(𝑗) (i.e. so that [𝑓(𝑖)] = [𝑓(𝑗)] in the quotient). When and only when this happens, we peek at the first 𝑁 + 2 values of 𝑓, where 𝑁 is some number larger than all the numbers in 𝑅 and 𝑆 (not counting reflexivity pairs).

It might be that within these 𝑁 + 2 values there are duplicates. If so we simply return their indices.

It might also be that there are no duplicates. But then there are at least two values 𝑓(𝑖) and 𝑓(𝑗) that exceed 𝑁. (After all, there are only 𝑁 numbers that are smaller than 𝑁.) We then add the pair (𝑓(𝑖), 𝑓(𝑗)) to 𝑅, preserving orthogonality with 𝑆, and return (𝑖,𝑗).

This way we can meet every challenge. Voilà, a pigeonhole-powered counterexample to the streamlessness conjecture.

But wait, how can we *modify* 𝑅 and 𝑆? Aren't mathematical objects set in stone, living out an eternal unchanging existence in platonic heaven?

To realize this fantasy as an actual countermodel, we turn to the technology described in the next post.