For finite collections of compact spaces, this is not very surprising and is easy to prove. The statement is in fact true for infinite collections of arbitrary size, depends heavily on the peculiar definition of the product topology in this case, and is equivalent to the axiom of choice.

Tychonoff's theorem is complex, and its proof is often approached in parts, proving helpful lemmas first. One approach to proving it (Munkres section 37) exploits an alternative formulation of compactness based on the finite intersection property. The two lemmas are shown follow:

**Lemma 1**: For any (the power set of`X`) with the finite intersection property (FIP), there is a maximal set`D`with FIP containing`A`. By "maximal" we mean that no collection satisfying FIP properly contains`D`.**Lemma 2**: If`D`is a maximal FIP-satisfying subset of , then any finite intersection of elements of`D`is contained in`D`, and any subset of`X`intersecting every element of`D`is also contained in`D`.

Once we have `D`, we can project it along each of the infinitely many dimensions to obtain FIP sets in the spaces forming the product. But we know these spaces are compact, and so we can choose a point in each space from the intersection of that space's projected `D` collection. These become the coordinates of an element `x` in the infinite product space.

Finally, it's possible to show that if one of the spaces in the product has a subbasis element containing that space's coordinate of `x`, then the "tube" formed by pulling that subbasis into the full product space with an inverse projection map will contain `x` and will also intersect every element of `D`. Lemma 2 then tells us that each of these tubes is in `D`. But tubes form a subbasis in the product topology, and so, also by Lemma 2, all basis elements containing `x` are in `D`. But then these basis elements intersect every element of `D`, and so `x` is a limit point of each element of `D`, and so is in the closure of each element of `D`.

It was mentioned above that Tychonoff's theorem is, in fact, *equivalent* to the axiom of choice (AC). This seems surprising at first, since AC is an entirely set-theoretic formulation, not mentioning topology at all. But in view of the complexity of the proof of Tychonoff's theorem, and that mathematics can be completely modeled in set theory (i.e. the category of sets is a topos), this is not altogether unexpected. This equivalence shows that the formulation of compactness in infinite product spaces is nonconstructive (also not altogether unexpected, since AC itself is equivalent to asserting whether or not an infinite product is empty!).

To prove that Tychonoff's theorem implies the axiom of choice, we establish that every infinite cartesian product of non-empty sets is nonempty. It is actually a more comprehensible proof than the above (probably because it does not involve Zorn's Lemma, which is quite opaque to most mathematicians as far as intuition is concerned!). The trickiest part of the proof is introducing the right topology. The right topology, as it turns out, is the cofinite topology with a small twist. It turns out that every set given this topology automatically becomes a compact space. Once we have this fact, Tychonoff's theorem can be applied; we then use the FIP definition of compactness (the FIP is sure convenient!). Anyway, to get to the proof itself (due to J.L. Kelley):

Let {`A _{i}`} be an indexed family of nonempty sets, for

Now the define cartesian product

Now here's the trick: we give each `X _{i}` the topology whose open sets are the cofinite subsets of