Why Box Products Can Fail to Be Compact

Two Topologies on the Same Product

For a finite product, the product and box topologies agree. For an infinite product, the box topology permits open sets that restrict every coordinate at once. That extra freedom matters: even when every factor is a compact interval, the box topology can fail to be compact.

We use \([0,1]\) with its usual subspace topology from \(\mathbb{R}\), and write \(X=[0,1]^{\mathbb{N}}\) for the set of all sequences \(x=(x_1,x_2,\ldots)\) with \(x_n\in[0,1]\). The product and box topologies below are two different topologies on this same set.

Definition: A box in \(X\) is a set \(\prod_{n\in\mathbb{N}}U_n\), where each \(U_n\) is open in \([0,1]\). These boxes form a basis for the box topology. A basis for the product topology consists of sets \(\prod_{n\in\mathbb{N}}U_n\) where each \(U_n\) is open in \([0,1]\) and \(U_n=[0,1]\) for all but finitely many \(n\).

Thus, to make a box-open neighborhood of a point, one may impose an open condition on every coordinate. A basic product-open neighborhood can impose conditions on only finitely many coordinates. In both definitions, the sets \(U_n\) are open in the factor \([0,1]\), not necessarily open as subsets of \(\mathbb{R}\).

Proposition: The product topology is contained in the box topology. In other words, every product-open set is box-open.

Proof. Each basic product-open set is a box: its coordinate sets \(U_n\) are open, and the unrestricted factors \([0,1]\) are open in themselves. Therefore every member of the product basis is box-open. Every product-open set is a union of members of that basis, and a union of box-open sets is box-open. \(\square\)

A Box-Open Cover with No Finite Subcover

For \(n\in\mathbb{N}\), let \(U_n\) require that the \(n\)th coordinate be less than \(1\), with no restrictions on the other coordinates. Also let \(V\) require that every coordinate be strictly greater than \(1/2\):

U_n=\prod_{k\in\mathbb{N}}W_k^{(n)},\qquad W_k^{(n)}= \begin{cases} [0,1),&k=n,\\ [0,1],&k\ne n, \end{cases} \qquad V=\prod_{k\in\mathbb{N}}(1/2,1].

These are box-open. Indeed, \([0,1)\) and \((1/2,1]\) are open in the subspace \([0,1]\): they are respectively \((-1,1)\cap[0,1]\) and \((1/2,2)\cap[0,1]\). The unrestricted factor \([0,1]\) is open in itself.

Worked Example: Checking That the Sets Cover

We verify that \(\{U_n:n\in\mathbb{N}\}\cup\{V\}\) covers \(X\). Let \(x=(x_1,x_2,\ldots)\in X\). If \(x_n<1\) for at least one index \(n\), then its \(n\)th coordinate belongs to \([0,1)\), and every other coordinate belongs to its unrestricted factor \([0,1]\). Hence \(x\in U_n\). If there is no such index, then \(x_n=1\) for every \(n\), since each coordinate is in \([0,1]\). In that case \(x\in V\), because \(1>1/2\) in every coordinate. These cases include every point of \(X\), so the sets cover \(X\).

Check: the all-ones sequence belongs to \(V\), since each of its coordinates satisfies \(1>1/2\). The sequence with first coordinate \(0\) and all remaining coordinates \(1\) belongs to \(U_1\), since \(0<1\).

Theorem: The box topology on \([0,1]^{\mathbb{N}}\) is not compact.

Proof. Compactness means that every open cover has a finite subcover. The family \(\{U_n:n\in\mathbb{N}\}\cup\{V\}\) is a box-open cover, as just verified. Consider any finite subfamily. First suppose it does not contain \(V\). It then consists of finitely many sets \(U_n\), and the all-ones sequence belongs to none of them: its \(n\)th coordinate is \(1\), which does not satisfy the strict condition \(x_n<1\). Thus this subfamily does not cover \(X\).

Now suppose the finite subfamily contains \(V\). Let \(S\subseteq\mathbb{N}\) be the finite set of indices \(n\) for which the subfamily contains \(U_n\). Choose \(m\in\mathbb{N}\setminus S\), which is possible because \(\mathbb{N}\) is infinite and \(S\) is finite. Define \(x\in X\) by setting \(x_i=1\) for \(i\in S\), \(x_m=0\), and \(x_i=1\) for every other index. For each \(n\in S\), we have \(x_n=1\), so \(x\notin U_n\) because \(U_n\) requires \(x_n<1\). Also \(x\notin V\), because its \(m\)th coordinate is \(0\), which does not satisfy \(x_m>1/2\). Therefore this finite subfamily misses \(x\). No finite subfamily covers \(X\), so the box topology is not compact. \(\square\)

Worked Example: A Finite Subfamily in Detail

Suppose a proposed finite subcover contains \(U_2\), \(U_5\), and \(V\), and no other sets. Its selected index set is \(S=\{2,5\}\). Choose \(m=1\), which is not in \(S\), and set \(x_1=0\) and \(x_i=1\) for every \(i\ne1\). The condition for \(U_2\) is \(x_2<1\), but \(x_2=1\), so \(x\notin U_2\). The condition for \(U_5\) is \(x_5<1\), but \(x_5=1\), so \(x\notin U_5\). Membership in \(V\) requires \(x_1>1/2\), but \(x_1=0\), so \(x\notin V\). This particular finite subfamily misses \(x\). The proof works for every finite selected index set \(S\), not only this choice.

Check: \(x\) is a point of \(X\), since \(0\in[0,1]\) and \(1\in[0,1]\). Each failed membership follows from the strict inequalities defining the sets.

Why This Cover Does Not Contradict Product Compactness

The set \(V\) above is box-open but is not product-open. This distinction is essential: the cover used to prove noncompactness is not an open cover for the product topology. The sets \(U_n\), which restrict only one coordinate each, are product-open; the difficulty is the infinitely many simultaneous restrictions defining \(V\).

Worked Example: The Set \(V\) Is Not Product-Open

Let \(p=(1,1,\ldots)\). We have \(p\in V\). A basic product-open neighborhood \(B\) of \(p\) restricts only finitely many coordinates, say those in a finite set \(F\subseteq\mathbb{N}\). Thus \(B=\prod_n O_n\), where each \(O_n\) is open in \([0,1]\), \(1\in O_n\), and \(O_n=[0,1]\) for \(n\notin F\). Choose \(m\notin F\), and define \(y\) by \(y_m=0\) and \(y_i=1\) for \(i\ne m\). The unrestricted \(m\)th factor contains \(0\), and every restricted factor contains \(1\); therefore \(y\in B\). But \(y\notin V\), because \(y_m=0\) and \(0\) is not strictly greater than \(1/2\).

Consequently, no basic product-open neighborhood of \(p\) is contained in \(V\). If \(V\) were product-open, its point \(p\) would have a basic product-open neighborhood contained in \(V\), by the definition of a basis. This contradiction proves that \(V\) is not product-open.

Check: the construction uses an index \(m\) outside the finite set \(F\). Such an index exists because \(F\) is finite whereas \(\mathbb{N}\) is infinite.

The product topology on \(X\) is compact. The earlier result known as Tychonoff’s theorem states that a product of compact spaces is compact when equipped with the product topology. Each factor \([0,1]\) is compact by the Heine–Borel theorem, so the hypotheses apply to the family \(([0,1])_{n\in\mathbb{N}}\). This is a statement about the product topology; it does not imply compactness for the finer box topology.

Finite Products and the Common Pitfall

Worked Example: Two Interval Factors

Consider \(Y=[0,1]\times[0,1]\). A box basis element has the form \(U_1\times U_2\), where both \(U_1\) and \(U_2\) are open in \([0,1]\). A product basis element has the same form: there are only two coordinates, so restricting both is still restricting only finitely many. Conversely, every product basis element is a box basis element. The two bases therefore generate the same topology on \(Y\).

The square \(Y\), with this topology, is compact: it is a closed and bounded subset of \(\mathbb{R}^2\), so compactness follows from Heine–Borel. Thus the finite-product case does not display the failure above. The construction depended on having an index outside every finite collection of restricted coordinates.

Check: for any \(U_1,U_2\) open in \([0,1]\), the set \(U_1\times U_2\) is allowed in both bases. There is no third or later coordinate on which the box basis could impose additional restrictions.

Common pitfall. A tempting but incorrect argument is: “Each factor \([0,1]\) is compact, so \([0,1]^{\mathbb{N}}\) is compact in either topology.” The conclusion does not follow in the box topology. In the product topology, Tychonoff’s theorem applies because the topology is the product topology; in the box topology, the explicit cover above has no finite subcover. Another common error is to call \(V=\prod_n(1/2,1]\) a basic product-open set because every factor is open. Product basic sets require all but finitely many factors to be the whole space. Here \((1/2,1]\ne[0,1]\) for every \(n\), so infinitely many coordinates are restricted.

The distinction is not that the factors have changed or that the underlying set has changed. Only the topology has changed. Box-open sets can test infinitely many coordinates simultaneously, and the cover exploits exactly that freedom. Compactness is sensitive to the choice of topology, so a compactness theorem for one topology cannot be transferred to another without checking its hypotheses.

Check Your Understanding

For each question, distinguish the basis requirements for the product and box topologies.

On \([0,1]^{\mathbb{N}}\), explain why each \(U_n\) in the cover is open in both topologies.
Why does the all-ones sequence fail to belong to every \(U_n\), and why does it belong to \(V\)?
Given any finite set \(S\subseteq\mathbb{N}\), construct a point missed by \(V\) and all sets \(U_n\) with \(n\in S\).
Explain why the proof that \(V\) is not product-open needs a coordinate outside the finite set of restricted coordinates.
Which topology on a product of compact spaces is covered by Tychonoff’s theorem?