Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 6 / 2

The Tychonoff theorem states that an arbitrary product of compact spaces, with the product topology, is compact; no Hausdorff hypothesis is necessary. We prove it using Zorn's lemma and the axiom of choice.

First, every proper filter on a set extends to an ultrafilter. Order its proper extensions by inclusion. The union of a chain is again a proper filter: any finite collection of its members occurs in one member of the chain, and the empty set occurs in none. Zorn's lemma gives a maximal proper filter $\mathcal{U}$. If $A\notin \mathcal{U}$, adjoining $A$ must destroy properness, so some $F\in \mathcal{U}$ has $F\cap A=\varnothing$, and hence $X\setminus A\in \mathcal{U}$. This proves the decision property of an ultrafilter.

Next prove the ultrafilter characterization of compactness. In a compact space the closed sets $\overline{F}$, for $F\in \mathcal{U}$, have the finite intersection property, so their intersection contains a point $x$. Every open neighbourhood $O$ of $x$ belongs to $\mathcal{U}$: otherwise its closed complement belongs to $\mathcal{U}$ and contains $x$, a contradiction. Thus the ultrafilter converges to $x$. Conversely, if an open cover has no finite subcover, the closed complements have the finite intersection property and generate a proper filter. Extend it to an ultrafilter. A limit point lies in some member $O$ of the cover, so both $O$ and its complement would belong to the ultrafilter. This is impossible. These arguments also prove the closed-set form of compactness directly by taking complements.

Now let $X={\prod }_{j\in J}{X}_j$. If a factor is empty, the product is empty and compact. Otherwise the axiom of choice makes $X$ nonempty. For an ultrafilter $\mathcal{U}$ on $X$, its coordinate imageis an ultrafilter. Compactness of $X_j$ supplies at least one limit $x_j$; use the axiom of choice to choose one for every $j$. A basic open neighbourhood of $x=(x_j{)}_j$ restricts finitely many coordinates. Each corresponding inverse image belongs to $\mathcal{U}$, so their finite intersection also does. Hence $\mathcal{U}$ converges to $x$ in the product topology. The ultrafilter criterion proves the theorem. The explicit choices and the maximal-filter argument are exactly the places where choice enters.

The Banach-Alaoglu theorem states that the closed unit ball of the continuous dual space $V^{\ast }$ of a real or complex normed vector space $V$ is compact in the weak-star topology $\sigma (V^{\ast },V)$. Completeness of $V$ is not required.

For each $v\in V$, let $D_v=\{z\in \mathbb{K}:|z|\le \Vert v\Vert \}$, a compact scalar disk or interval. By the Tychonoff theorem, $P={\prod }_{v\in V}{D}_v$ is compact. Inside $P$, impose the equationsEach equation defines a closed subset, since it concerns continuous coordinate projections into a Hausdorff scalar field. Their intersection $K$ is therefore compact. A point in $K$ defines a linear functional $\ell (v)=z_v$, and the coordinate bounds give $|\ell (v)|\le \Vert v\Vert$. Conversely, every $\ell \in V^{\ast }$ with $\Vert \ell \Vert \le 1$ gives such a point. Thus $K$ is exactly the image of the dual unit ball under $\ell \mapsto (\ell (v){)}_{v\in V}$. The induced product topology is precisely pointwise convergence on $V$, which is the weak-star topology. This proves

This proof uses neither the axiom of choice nor a preliminary choice of points in all the given sets. For each nonempty $A_j$, form a disjoint union $X_j=A_j\bigsqcup \{{\infty }_j\}$ with a distinguished added point, and give it the topologyThe original set $A_j$ has the indiscrete subspace topology, and is clopen. Any open cover of $X_j$ either contains $X_j$, or contains both $A_j$ and $\{{\infty }_j\}$, so it has a subcover of at most two members. Thus $X_j$ is compact by a proof requiring no choices. No Hausdorff restriction is imposed in this question, and these generally non-Hausdorff spaces are allowed.

The product $X={\prod }_{j\ge 1}{X}_j$ has the explicitly given all-infinity point, so it is nonempty without choice. By the assumed countable-product compactness it is compact. For each $j$ letThe coordinate projection is continuous and $A_j$ is closed in $X_j$, so $F_j$ is closed. Any finite intersection of the $F_j$ is nonempty: pick an element of each of those finitely many nonempty sets and put ${\infty }_k$ in every other coordinate. Finite choice follows by induction in ordinary set theory without assuming the axiom of choice.

Compactness and the finite intersection property now give ${\bigcap }_{j\ge 1}{F}_j\ne \varnothing$. Choose one point $x$ from this single nonempty set. The function $f(j)=x_j$ satisfies $f(j)\in A_j$ for every $j$, proving the axiom of countable choice.

The implication holds in choice-free set theory: countable product compactness implies countable choice. The printed lower index zero is inconsistent with the stated positive-integer index set; indexing from one merely relabels this argument and changes no conclusion.