Why Infinite Products Raise a Compactness Question
For a finite product, it is natural to expect compactness to be governed by the compactness of its factors. An infinite product raises a sharper question: what does a neighborhood look like when there are infinitely many coordinates, and can compactness still be controlled coordinate by coordinate? The answer begins with the product topology. Its basic open sets place restrictions on only finitely many coordinates at a time.
Let \(J\) be an index set, and let \((X_j,\mathcal{T}_j)\) be a topological space for each \(j\in J\). The underlying set of the product is \(\prod_{j\in J}X_j\), whose points are functions \(x\) with \(x_j\in X_j\) for each \(j\). The product topology is defined by the basis below. If \(J\) is empty, the product is a one-point space; if any factor is empty, the product is empty.
Thus a basic neighborhood of a point can narrow down finitely many coordinates and leave every other coordinate unrestricted. This does not mean that an open set must involve only finitely many coordinates: an open set may be a union of arbitrarily many basic open sets. Rather, each point in an open set has at least one basic neighborhood around it that makes only finitely many coordinate restrictions.
Proof. First, every point \(x\) of the product belongs to the basic open set \(\prod_{j\in J}X_j\), so the basic sets cover the product. Now take two basic open sets \(B=\prod_{j\in J}U_j\) and \(C=\prod_{j\in J}V_j\). Their intersection is \(\prod_{j\in J}(U_j\cap V_j)\). Each coordinate intersection is open, and outside the union of the two finite sets of restricted coordinates it equals \(X_j\). The union of two finite sets is finite, so \(B\cap C\) is basic open. These covering and intersection properties are the basis conditions. \(\square\)
Reading Basic Neighborhoods
Worked Example: A Cylinder in a Product of Finite Discrete Spaces
For each \(j\in\mathbb{N}\), let \(X_j=\{0,1,2\}\) with the discrete topology, in which every subset is open. In \(X=\prod_{j\in\mathbb{N}}X_j\), consider \(B=\{x\in X:x_2=1,\ x_5=0\}\).
For coordinate \(2\), take \(U_2=\{1\}\). For coordinate \(5\), take \(U_5=\{0\}\). At every other coordinate \(j\), take \(U_j=X_j\). These are open sets in their respective factors, and only two coordinates are restricted. Therefore \(B=\prod_j U_j\) is basic open in the product topology.
For example, the point \(x\) with \(x_j=0\) for every \(j\) does not belong to \(B\), because \(x_2=0\ne1\). The point \(y\) with \(y_2=1\), \(y_5=0\), and \(y_j=2\) at every other coordinate does belong to \(B\): its second coordinate is \(1\) and its fifth coordinate is \(0\). No condition has been placed on its other coordinates.
Check: the defining coordinate tests for \(y\) give \(y_2=1\) and \(y_5=0\), so \(y\in B\). For \(x\), \(x_2=0\), so \(x\notin B\), regardless of its fifth coordinate.
Worked Example: A Neighborhood in a Product of Intervals
Give each copy of \([0,1]\) its subspace topology from the standard topology on \(\mathbb{R}\), and set \(X=\prod_{j\in\mathbb{N}}[0,1]\). Define \(B=\{x\in X:x_2\in(1/4,3/4),\ x_7\in[0,1/3)\}\).
The set \((1/4,3/4)\) is open in \([0,1]\), since it is the intersection of \([0,1]\) with the open interval \((1/4,3/4)\) in \(\mathbb{R}\). The set \([0,1/3)\) is also open in \([0,1]\), since it is the intersection of \([0,1]\) with the open interval \((-1,1/3)\). Taking these sets at coordinates \(2\) and \(7\), and the full interval at all other coordinates, shows that \(B\) is basic open.
Let \(x_j=1/2\) for every \(j\). The second coordinate satisfies \(1/4<1/2<3/4\), and the seventh satisfies \(0\le 1/2\) but not \(1/2<1/3\). Thus \(x\notin B\). Instead, let \(y_2=1/2\), \(y_7=1/4\), and \(y_j=1/2\) at all other coordinates. Then \(1/4<y_2<3/4\), and \(0\le y_7=1/4<1/3\), so \(y\in B\).
Check: the restrictions concern only coordinates \(2\) and \(7\); the other entries of \(y\) need only belong to \([0,1]\), which they do.
Product Topology Is Not Box Topology
A tempting alternative is to allow a separate open restriction at every coordinate in a basic set. That defines the box topology, not the product topology. For an infinite product the difference matters: an all-coordinate restriction can be box-open without being product-open.
Worked Example: An All-Coordinate Restriction
Again take \(X=\prod_{j\in\mathbb{N}}[0,1]\), and define \(W=\prod_{j\in\mathbb{N}}[0,1/2)\). Each factor \([0,1/2)\) is open in \([0,1]\), so \(W\) is open in the box topology. We show that \(W\) is not open in the product topology.
Let \(o\) be the point with \(o_j=0\) for every \(j\). It belongs to \(W\). If \(W\) were product-open, the basis property would give a basic product-open set \(B=\prod_j U_j\) with \(o\in B\subseteq W\). Only finitely many coordinates of \(B\) can be restricted. Choose an index \(k\) that is not among them; then \(U_k=[0,1]\). Define \(z_k=3/4\) and \(z_j=0\) for every \(j\ne k\). Since \(z_k\in U_k\), and every other coordinate is \(0\in U_j\), we have \(z\in B\). But \(z_k=3/4\notin[0,1/2)\), so \(z\notin W\), contradicting \(B\subseteq W\). Therefore \(W\) is not open in the product topology.
Check: \(3/4\ge1/2\), so it fails the strict condition \(z_k<1/2\) required for membership in \(W\). The contradiction used a coordinate left unrestricted by the basic product neighborhood.
Common pitfall. One might write “each coordinate restriction is open, so their product is open” and conclude that \(W\) is product-open. That argument silently uses the box topology. In the product topology, a basic open set may restrict only finitely many coordinates. The correction is to check the finite-restriction condition, not merely the openness of each factor set.
A Necessary Condition for Product Compactness
Compactness of a product forces compactness of every factor, provided the other factors make each coordinate projection onto. The relevant map is the coordinate projection \(\pi_k:\prod_j X_j\to X_k\), defined by \(\pi_k(x)=x_k\).
Proof. Fix \(k\in J\). The map \(\pi_k\) is continuous: for any open \(U\subseteq X_k\), its preimage is the basic open set \(\pi_k^{-1}(U)=\{x:x_k\in U\}\), which restricts only coordinate \(k\). The map is surjective because every other factor is nonempty: for any \(a\in X_k\), choose a point in each other factor and form a product point with \(k\)th coordinate \(a\). (For arbitrary index sets, this selection uses the axiom of choice.)
It remains to verify that a continuous image of a compact space is compact. Let \(\{V_\alpha\}\) be an open cover of \(\pi_k(\prod_jX_j)=X_k\). Continuity makes the sets \(\pi_k^{-1}(V_\alpha)\) open, and surjectivity makes them cover the product. Compactness supplies finitely many of these preimages that cover the product. Their corresponding finitely many sets \(V_\alpha\) cover \(X_k\), again by surjectivity. Thus every open cover of \(X_k\) has a finite subcover, so \(X_k\) is compact. \(\square\)
The nonempty-factor condition prevents a degenerate exception: if one factor is empty, the whole product is empty and hence compact, even if another factor is noncompact. In that case the projection onto the other factor is not surjective, so the proof above does not apply.
Worked Example: A Noncompact Factor Rules Out Compactness
Give \(\mathbb{R}\) its standard topology and let \(Y=\prod_{j\in\mathbb{N}}[0,1]\), with the product topology from the interval topologies. Consider \(X=\mathbb{R}\times Y\), also with the product topology. For each \(n\in\mathbb{N}\), define \(V_n=(-n,n)\times Y\).
Each \(V_n\) is open: it restricts the \(\mathbb{R}\)-coordinate to the open interval \((-n,n)\) and leaves the \(Y\)-coordinate unrestricted. The family covers \(X\), since for any \(t\in\mathbb{R}\), there is a positive integer \(n>|t|\), and then \((t,y)\in V_n\) for every \(y\in Y\).
No finite subfamily covers \(X\). Given finitely many indices, let \(N\) be their largest. The intervals are nested, so the union of the corresponding sets is contained in \(V_N=(-N,N)\times Y\). The point \((N,y)\), for any \(y\in Y\), is not in \(V_N\), because the interval has strict endpoints and \(N\notin(-N,N)\). Hence the cover has no finite subcover, and \(X\) is not compact.
Check: \(N\) is an endpoint, not an interior point, of \((-N,N)\). Thus \((N,y)\notin V_N\), verifying the failure of every finite subcover directly.
What These Tests Suggest—and What They Do Not Prove
The projection result supplies a useful first test: if a product with nonempty factors is compact, no coordinate factor can be noncompact. It does not, by itself, prove that compactness of all factors is sufficient. That converse is the content of the Tychonoff theorem: a product of compact spaces is compact in the product topology. Its proof for arbitrary products uses the axiom of choice and is developed later in the material on filters and ultrafilters.
So products of finite discrete spaces and products of compact intervals are plausible compactness candidates, while a product with a noncompact factor is ruled out by the projection test. The finite-coordinate basis explains why this plausible claim is not merely the finite-product argument repeated: neighborhoods inspect only finitely many coordinates even when the product has infinitely many. The box topology changes that local structure and is not covered by the product compactness theorem stated above.
| Question | What the product topology gives | What follows for compactness |
|---|---|---|
| What is a basic open set? | Open restrictions in finitely many coordinates; all other factors are unrestricted. | Every open neighborhood contains such a basic neighborhood. |
| Can a product be compact if a factor is noncompact? | Coordinate projections are continuous and, when all factors are nonempty, surjective. | No: a compact product would have compact coordinate images. |
| If every factor is compact, is the product compact? | The finite-coordinate basis makes this plausible. | The general answer is Tychonoff's theorem, not a consequence of the projection test alone. |
Check Your Understanding
Use the finite-coordinate basis and the compactness tests above to answer the following.
- In \(\prod_{j\in\mathbb{N}}\{0,1\}\) with discrete factors and the product topology, describe a basic open set that requires \(x_3=0\) and \(x_8=1\).
- Explain why \(\prod_{j\in\mathbb{N}}[0,1/2)\) is not open in the product topology on \(\prod_{j\in\mathbb{N}}[0,1]\).
- Suppose all factors are nonempty and one factor is noncompact. What does the coordinate projection imply about compactness of the product?
- Why does the projection argument for a compact product fail when some factor is empty?
- State the difference between the product and box topologies in terms of how many coordinates a basic open set may restrict.