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== Tychonoff's theorem and the axiom of choice == All of the above proofs use the [[axiom of choice]] (AC) in some way. For instance, the third proof uses that every filter is contained in an ultrafilter (i.e., a maximal filter), and this is seen by invoking [[Zorn's lemma]]. Zorn's lemma is also used to prove Kelley's theorem, that every net has a universal subnet. In fact these uses of AC are essential: in 1950 Kelley proved that Tychonoff's theorem implies the axiom of choice in [[Zermelo–Fraenkel set theory|'''ZF''']]. Note that one formulation of AC is that the Cartesian product of a family of nonempty sets is nonempty; but since the empty set is most certainly compact, the proof cannot proceed along such straightforward lines. Thus Tychonoff's theorem joins several other basic theorems (e.g. that every vector space has a basis) in being ''equivalent'' to AC. On the other hand, the statement that every filter is contained in an ultrafilter does not imply AC. Indeed, it is not hard to see that it is equivalent to the [[Boolean prime ideal theorem]] (BPI), a well-known intermediate point between the axioms of Zermelo-Fraenkel set theory (ZF) and the ZF theory augmented by the axiom of choice (ZFC). A first glance at the second proof of Tychnoff may suggest that the proof uses no more than (BPI), in contradiction to the above. However, the spaces in which every convergent filter has a unique limit are precisely the Hausdorff spaces. In general we must select, for each element of the index set, an element of the nonempty set of limits of the projected ultrafilter base, and of course this uses AC. However, it also shows that the compactness of the product of compact Hausdorff spaces can be proved using (BPI), and in fact the converse also holds. Studying the ''strength'' of Tychonoff's theorem for various restricted classes of spaces is an active area in [[set-theoretic topology]]. The analogue of Tychonoff's theorem in [[pointless topology]] does not require any form of the axiom of choice.
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