Compactness of a Product Forces Compactness of Its Factors

From a Compact Product to Its Factors

A product space carries information about each of its coordinates, but recovering a whole factor from the product requires that the other factor provide at least one point to pair with every point in the factor of interest. Coordinate projections make this precise: they are continuous, and their images depend on whether the opposite factor is empty. We will use these facts together with the continuous-image theorem to identify exactly when compactness passes from a product to a factor.

Throughout, \(X\) and \(Y\) are topological spaces with topologies \(\mathcal{T}_X\) and \(\mathcal{T}_Y\). Their product \(X\times Y\) has the product topology, whose basis consists of the sets \(U\times V\), where \(U\) is open in \(X\) and \(V\) is open in \(Y\). A space is compact if every open cover of it has a finite subcover. An open cover of a space is a family of open subsets whose union is the whole space.

The theorem applies to a map’s image, not automatically to its entire target. For a projection to show that a factor is compact, we therefore need to know when the projection’s image is the whole factor.

Coordinate Projections and Their Images

Proof. Let \(U\) be open in \(X\). Then \(\pi_X^{-1}(U)=U\times Y\), which is open in the product topology because \(U\) and \(Y\) are open in their respective spaces. Thus preimages of open sets under \(\pi_X\) are open, so \(\pi_X\) is continuous. For an open set \(V\) in \(Y\), \(\pi_Y^{-1}(V)=X\times V\) is open for the same reason, so \(\pi_Y\) is continuous. If \(Y\ne\varnothing\), choose \(y_0\in Y\). For every \(x\in X\), \((x,y_0)\in X\times Y\) and \(\pi_X(x,y_0)=x\), so \(\pi_X(X\times Y)=X\). If \(Y=\varnothing\), there are no ordered pairs in \(X\times Y\), so the image is empty. For \(\pi_Y\), if \(X\ne\varnothing\), choose \(x_0\in X\); then each \(y\in Y\) is the image of \((x_0,y)\), giving \(\pi_Y(X\times Y)=Y\). If \(X=\varnothing\), the domain is empty and the image is empty. These arguments also cover an empty target: the image of a map with empty codomain is empty. \(\square\)

The image formulas are the key point. If the factor opposite the projection is nonempty, the projection reaches every point of the target. If that factor is empty, the product has no points, so the projection has empty image—even if its target is nonempty.

The Compactness Implication

Proof. Suppose \(X\times Y\) is compact and \(Y\ne\varnothing\). The projection \(\pi_X\) is continuous, so the continuous-image theorem implies that \(\pi_X(X\times Y)\) is compact. The image formula gives \(\pi_X(X\times Y)=X\), and hence \(X\) is compact. Now suppose \(X\times Y\) is compact and \(X\ne\varnothing\). The continuous map \(\pi_Y\) has image \(Y\), so the continuous-image theorem gives that \(Y\) is compact. \(\square\)

Each nonemptiness assumption belongs to the factor being projected away from: to infer compactness of \(X\), we need \(Y\ne\varnothing\); to infer compactness of \(Y\), we need \(X\ne\varnothing\). These are not assumptions that the factor being concluded compact must already be nonempty.

If both factors are empty, both are compact: the empty space is compact because the empty subfamily is a finite subcover of any open cover of it. If exactly one factor is empty, however, the product is empty and therefore compact, while the nonempty factor need not be compact. The following examples make these cases explicit.

A Common Pitfall: Incorrect “If and Only If” Claims

It is tempting to compress the projection image statement into “\(\pi_X(X\times Y)=X\) if and only if \(Y\ne\varnothing\).” This is false: when \(X=Y=\varnothing\), the image and target are both empty, so they are equal even though \(Y\) is not nonempty. The symmetric claim about \(\pi_Y\) also fails: if \(X\ne\varnothing\) and \(Y=\varnothing\), then \(\pi_Y(X\times Y)=\varnothing=Y\), despite \(X\ne\varnothing\).

The correct equivalences, obtained directly from the image formulas, are \(\pi_X(X\times Y)=X\) if and only if \(Y\ne\varnothing\) or \(X=\varnothing\), and \(\pi_Y(X\times Y)=Y\) if and only if \(X\ne\varnothing\) or \(Y=\varnothing\). For the first equivalence, if \(Y\ne\varnothing\), the image is \(X\); if \(X=\varnothing\), both image and target are empty. Conversely, if \(Y=\varnothing\) and \(X\ne\varnothing\), the image is empty and differs from \(X\). The second equivalence follows by the same case check: if \(X\ne\varnothing\), the image is \(Y\); if \(Y=\varnothing\), both are empty; and if \(X=\varnothing\), \(Y\ne\varnothing\), the image differs from \(Y\).

The compactness theorem uses the sufficient hypotheses \(Y\ne\varnothing\) and \(X\ne\varnothing\), respectively, not these more complicated equivalences. Those hypotheses guarantee that the relevant projection is onto the factor whose compactness is sought.

Why the Projection Argument Matters

The argument is useful because it reduces a claim about a product to a familiar theorem about continuous images. The proof has three parts: identify a continuous map, determine its image, and apply preservation of compactness to that image. Omitting the image check is a common source of errors: a continuous map from a compact space guarantees compactness of the image, but a target that is larger than the image need not be compact.

The nonempty condition is not a technical detail that can be discarded after the proof. An empty factor makes the whole product empty regardless of the other factor’s topology or compactness. Thus compactness of the product alone cannot detect whether a possibly noncompact space was paired with the empty space. When the opposite factor is nonempty, every point in the desired factor appears as a coordinate of some product point, and the projection transfers compactness to the whole factor.