Originally published at 狗和留美者不得入内. You can comment here or there.

We defined a base of a topological space in Definition 4 of [1]. In short, a base via unions generates the topological space. Since the intersection of finitely many open sets is also an open set (this follows directly from one of the neighborhood axioms), one naturally also defines a set that generates the topological space per arbitrary unions of finite intersections. We thus have

**Definition 1** A *subbase* of a topological space is a collection of its open sets, , such that for every open set , there exists a finite subset of , such that .

**Proposition 1** A base of a topological space is also a subbase of it.

*Proof*: Trivial.

Since Tychonoff’s theorem involves product spaces, we shall also define them.

**Definition 2** Let and be topological spaces. is *continuous* iff for every sequence in that converges to some , the sequence also converges to . Equivalently, if is open in , then is open in .

We leave the proof of the equivalence in Definition 2 to the reader.

**Definition 3** Let be a collection of topological spaces indexed by for where is an arbitrary set. We take to be the Cartesian product of all the topological spaces in the aforementioned collection. We let be the *canonical projection* to the component corresponding to index , which we require to be continuous for all .

**Proposition 2** Every topology on must contain, for all open in for all , .

*Proof*: This must hold if the canonical projections are to be continuous.

**Definition 4** We define the *product topology* or *Tychonoff topology* on to be the coarsest topology on it such that canonical projections are continuous. This is the topology generated by via unions of finite intersections. In other words, this is a subbase of the product topology. One can also say that the open sets on are the subsets of it such that all but a finite number of the canonical projections across all , when applied to the subset in question, yield (the codomain of ), and that those which do not yield the codomain, are all open (in the topological space defined on the codomain).

**Tychonoff’s theorem (Theorem 1)** The topological space on the Cartesian product of any collection of compact topological spaces with the product topology is compact.

*Proof*: Definition 4 tells us that is a subbase for the product topology. Take any cover of the product space by elements of this subbase. We assume does not contain , in which case Tychonoff’s theorem holds trivially. With the exception of , every element of this subbase has exactly one for which the projection onto the component is not the . This induces a map from the subbase to . Thus, we can partition into the disjoint union , where is the preimage of of the aforementioned map. Suppose that no covers the space. Then for each we pick an not such that , which gives a point in not in any of element of , which would imply that is not a cover, a contradiction.

We note that covers the space for some and , where every is equal to for some open in . We see that is an open cover of , which, being compact, has a finite subcover, which we denote as . Then is a finite subcover of , which covers the product space. This shows that if we can prove that a topological space such that every cover from a subbase is compact, we have proven Tychonoff’s theorem. The aforementioned proposition is called the Alexander subbase theorem, named after American mathematician James Waddell Alexander II, which we shall prove next.

**Alexander’s subbase theorem (Theorem 2)** If every open cover from a subbase of a topological space has a finite subcover, then the topological space is compact.

*Proof*: We denote the cover by elements of the subbase as . We also remark that any open cover is a collection of sets and that there is a partial order on open covers via collection inclusion. Assume the hypothesis of the theorem, and then assume by contradiction, there there are open covers which do not have finite subcovers. Take all the collection of all such open covers. Let for be any collection of covers without finite subcovers and a total order such that whenever for , . Let . If had a finite subcover by sets , then we can find such that , in which case the maximum of per the total order defined on , which we denote as , is such that has a finite subcover, a contradiction. This means that every chain in open covers without finite subcovers is bounded above by some open cover without a finite subcover. This enables us to apply Zorn’s lemma (proven in [2]) to derive some open cover with finite subcover that is not a proper subset of any other such open cover, which we call . By our hypothesis on the subbase, is not a cover. Thus, we can take an in our topological space not contained by . Any containing is not in . For any where , must contain a finite subcover, which we denote as . , which also means that Let . By the definition of subbase, there exists that such and . Let be a finite subcover for the cover for all . Then, would be a finite subcover of , a contradiction. This completes our proof.

I shall now explain a bit of the intuition behind the proof of Tychonoff’s theorem using the Alexander subbase theorem. I believe it should be more or less immediately for anyone who has studied boolean algebra deeply.

A difficulty some students might have with point set topology is the existence of not only countable unions or products but also of uncountable ones, as in the general Tychonoff’s theorem. We note how here de Morgan’s laws still apply. A element’s not being in an countable union of sets means it is not in any of the uncountably many unioned sets. One, after putting the NOT or set complement operator inside into every set unioned also switches the union to intersection. In the proof of Tychonoff’s theorem, this immediately tells us that the proposition of not being a cover evaluates to true iff a conjunction of nots of set inclusion evaluates to true, which can only be violated if no covers the space. This is basically saying that the “true for all” operator applied to a collection of boolean values evaluates to true iff all of the elements of the collection are true. From this, we can see that in general, if we want to prove something given the hypothesis of membership in a union of sets, it suffices to prove anything that implies non-membership in all the unioned sets of the element in question.

Inclusion of an element in a Cartesian product of sets a logical conjunction operator of the inclusion of the projections into their corresponding sets. lets us isolate to a component of the product space to give us a covered guarantee based only on the value of one component. If the value of the that component does not lie in the guaranteed range of values, it is covered iff the value of some other component lies in the guaranteed range of values for that other component. If for some , that guarantee obviously always exists. If that doesn’t hold for any , then there exists an uncovered point.

**References**