2026-04-18 16:16 UTC
ah, I think I have a counterexample for "if boundary is tame, then occupancy must be monotone or antitone in each variable"
This shape is not monotone or antitone in x.
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@jcreed@mastodon.social 2026-04-18 16:25
So here's a restatement of the question that I still don't know the answer to: Let f : 𝔹ⁿ → 𝔹 be a boolean function. Turn this into a subset of ℝⁿ by saying S = { v ∈ ℝⁿ | f(v₁ ≥ 0, …, vₙ ≥ 0) } Is there a nice purely combinatorial condition on f that is equivalent to "the boundary ∂S is homeomorphic to ℝⁿ⁻¹"?