Does anyone know if axiom of choice is nessesary in proving Tychonoff's theorem in a Hausdorff space? Thanks!
2026-04-04 12:00:14.1775304014
Tychonoff's theorem and axiom of choice in Hausdorff spaces
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No. The axiom of choice is not fully needed for this result, but you do need some fragment of the axiom of choice.
In fact, Tychonoff theorem for Hausdorff spaces is equivalent to the Boolean Prime Ideal theorem, which in turn is equivalent to the ultrafilter theorem, as well the completeness theorem and the compactness theorem (of first-order logic).
You can find the details in Herrlich's The Axiom of Choice (Theorem 4.70).