Infinite Parity Continuity Obstruction

Abstract

Total parity on finite-support Boolean configurations has no continuous completion on the full countable product.

Theorem 1.1 (Finite-support total parity has no continuous completion).

$$\neg \left(\exists extension\in \left(\mathbb{N}\to Bool\right)\to Bool,\mathrm{Continuous}\left(extension\right)\wedge \left(\forall support\in \mathrm{Finset}\left(\mathbb{N}\right),extension\left(\mathrm{readout}\left(support\right)\right)=\mathrm{decide}\left(\mathrm{Odd}\left(\mathrm{card}\left(support\right)\right)\right)\right)\right)$$

Proof. Machine-checked in Lean as D5/S3/Observer/Completion/InfiniteParityContinuityObstruction.finite_support_parity_has_no_continuous_completion (✓ std3). ∎

Source. Repository-derived.

Commentary.

A finite set of natural-number coordinates represents a finite prime configuration through the canonical readout map. Its total parity is the Boolean decision of odd support cardinality.

The initial-segment configurations converge coordinatewise to the all-active path. Continuity into the discrete Boolean space would make their parity eventually constant, while consecutive even and odd prefix lengths always give different values.

References

• Truth anchor: D5/S3/Observer/Completion/InfiniteParityContinuityObstruction.finite_support_parity_has_no_continuous_completion
• Dependency: D5/S3/Observer/ProbabilisticClosure/FiniteMarginalGlobalReadoutContrast