Compactness and Nonemptiness Are Different Questions
For a family of topological spaces, Tychonoff's theorem concerns whether the product is compact. A separate question is whether the product has any points at all. The distinction matters at the boundary cases: a product with an empty factor has no points, while a product with no factors has exactly one point. Both products are compact. Thus compactness, which is a condition on open covers, does not by itself guarantee that a space is nonempty.
We will interpret a product set as a set of choice functions. This definition makes the cases of an empty index set and an empty factor precise, and also identifies where the axiom of choice enters. Whenever we discuss a topological product, it has the product topology.
For topological spaces \((X_i,\mathcal{T}_i)\), the product topology has as a basis the sets restricting only finitely many coordinates: each basic set has the form \(\prod_{i\in I}U_i\), where every \(U_i\) is open in \(X_i\) and \(U_i=X_i\) except for finitely many indices. The topology on the empty set is \(\{\varnothing\}\); on a singleton it is \(\{\varnothing,X\}\). These facts will determine the topologies in the boundary cases below.
When Does a Product Have a Point?
Proof. By the definition of the product, each of its elements is exactly a function \(f\) defined on \(I\) such that \(f(i)\in X_i\) for every \(i\in I\). Such a function is a choice function, so the product contains a point if and only if a choice function exists. If \(X_j=\varnothing\), no value \(f(j)\) can belong to \(X_j\), so no such function exists. If \(I=\varnothing\), the empty function meets the condition at every index, because there are no indices to check. It is the only function with empty domain, so the product has exactly one element. \(\square\)
The last case is sometimes called the empty product. Its single point is itself the empty function; the product is the set containing that function. This distinction is important: the empty product is not the empty set. In contrast, if there is at least one index and a factor at that index is empty, there can be no function selecting a value from every factor.
Worked Example: The Product with No Factors
Let \(I=\varnothing\). Compute \(\prod_{i\in I}X_i\), even though there are no spaces \(X_i\) in the family.
A product element would have to be a function with domain \(\varnothing\), with no coordinate conditions to satisfy. There is exactly one such function, namely the empty function \(\varnothing\). Therefore \[ \prod_{i\in\varnothing}X_i=\{\varnothing\}. \] The product is a one-point space. Its product topology contains \(\varnothing\) and the whole product \(\{\varnothing\}\), so its open sets are precisely \(\varnothing\) and \(\{\varnothing\}\).
To check compactness, consider any open cover of this one-point space. At least one member of the cover contains its point \(\varnothing\). That single member is a finite subcover. Hence the empty product is compact and nonempty.
Worked Example: A Product with an Empty Factor
Let \(I=\{1,2\}\). Give \(X_1=\{a,b\}\) the discrete topology, and give \(X_2=\varnothing\) its unique topology \(\{\varnothing\}\). Determine the product and its compactness.
An element of \(X_1\times X_2\) would have to be a function \(f\) with \(f(1)\in\{a,b\}\) and \(f(2)\in\varnothing\). The second condition cannot be met. Thus \[ X_1\times X_2=\varnothing. \] The product topology on this set is its unique topology, \(\{\varnothing\}\).
The empty space is compact. Indeed, an open cover of \(\varnothing\) has the empty subfamily as a finite subcover: the union of no sets is \(\varnothing\), which already covers the empty space. Therefore this product is compact but has no points. Notice that \(X_1\) is nonempty and compact, and \(X_2\) is also compact: every open cover of the empty space has the finite subcover just described.
Check: the only point in the proposed product would need a second coordinate in \(X_2=\varnothing\), and no such coordinate exists. So the product is empty, not a one-point space.
Proof. For the empty space, the empty subfamily of any open cover is finite and has union equal to the whole space, so it is a finite subcover. For a one-point space \(\{p\}\), any open cover must contain some open set \(U\) with \(p\in U\). The single set \(U\) covers the space, so it is a finite subcover. \(\square\)
What Tychonoff's Theorem Does—and Does Not—Say
Tychonoff's theorem states that, in the usual set-theoretic setting with the axiom of choice, a product of compact spaces with the product topology is compact. The compactness conclusion is consistent with either of the two boundary cases: a product can be empty because a factor is empty, or it can be a singleton because there are no factors. The theorem's compactness conclusion is not an existence claim for a point.
For clarity, the axiom of choice is the assertion that every family of nonempty sets has a choice function. By the definition above, this is precisely the assertion that \(\prod_{i\in I}X_i\ne\varnothing\) whenever every \(X_i\ne\varnothing\), for every index set \(I\). In particular, when the index set is arbitrary, the claim that nonempty factors always have a nonempty product is the axiom of choice, not a consequence of compactness.
Worked Example: A Finite Product with Explicit Choices
Let \(I=\{1,2\}\), where \(X_1=\{a,b\}\) and \(X_2=\{0,1\}\), and give both factors the discrete topology. List the product points and verify compactness directly.
A product point is a function choosing one of \(a,b\) at index \(1\) and one of \(0,1\) at index \(2\). The four possible functions are \[ f_{a,0}=(a,0),\qquad f_{a,1}=(a,1),\qquad f_{b,0}=(b,0),\qquad f_{b,1}=(b,1). \] Thus \(X_1\times X_2\) has exactly four points. In the product topology, each singleton is open: for example, \(\{a\}\times\{0\}\) is a basic open set containing only \((a,0)\), because both coordinate singletons are open in their discrete topologies. The same construction gives an open singleton around each of the other three points. Every subset is therefore open, and the product topology is discrete.
To verify compactness, take any open cover of the product. For each of its four points, select one cover member containing that point. The at most four selected members cover all four points, so they form a finite subcover. The product is nonempty because the displayed functions give its points; its compactness follows directly from its finite size.
Check: the list has two choices for the first coordinate and two for the second, giving \(2\cdot2=4\) functions, exactly the four listed. Each listed function satisfies both coordinate membership conditions.
This finite example also indicates why finite products are different from arbitrary products in the nonemptiness question. For a finite family of nonempty sets, one can successively choose an element from each factor using only finitely many choices. For an arbitrary family, selecting one element from every factor is exactly the assertion made by the axiom of choice. Compactness is a separate property and does not supply the missing selections.
This theorem is the standard Tychonoff result established by the earlier material on ultrafilters and products. Here its role is to clarify the scope of the conclusion, not to supply a choice function. If all factors are nonempty, the product is nonempty exactly when a choice function exists. If any factor is empty, compactness may still hold, as the second worked example shows. If there are no factors, the product is the singleton from the first example.
A Common Mistake and Its Correction
Common pitfall. One might reason, “There are no coordinates in the empty product, so there are no product points,” and write \(\prod_{i\in\varnothing}X_i=\varnothing\). This confuses the empty index set with an empty factor. A product point is a function satisfying a condition at every index. When there are no indices, the empty function satisfies every such condition: there is no coordinate at which it could fail. Consequently the product is \(\{\varnothing\}\), not \(\varnothing\).
The converse error is also possible: treating the presence of a nonempty factor as enough to make a product nonempty. For \(I=\{1,2\}\), \(X_1=\{a,b\}\), and \(X_2=\varnothing\), there are no functions that choose a value in both factors. The product is empty even though \(X_1\) is nonempty. To test nonemptiness, check the whole family: an element must choose a value at every index.
| Situation | Product as a set | Compact? | Nonempty? |
|---|---|---|---|
| No indices: \(I=\varnothing\) | The singleton containing the empty function | Yes | Yes |
| At least one index with an empty factor | The empty set | Yes | No |
| All factors nonempty, with a choice function | Contains that function as a point | Yes if all factors are compact, by Tychonoff | Yes |
| All factors nonempty, for an arbitrary family | Nonempty exactly when a choice function exists | Compact if the factors are compact, by Tychonoff with choice | Not guaranteed without the choice assertion |
Check Your Understanding
For each question, keep the set-theoretic question of having a point separate from the topological question of being compact.
- Describe the product set and product topology when the index set is empty. Why is it compact?
- Let \(I=\{1,2,3\}\), with \(X_1=\{u\}\), \(X_2=\{v,w\}\), and \(X_3=\varnothing\). Is the product empty? Explain by checking the coordinate conditions.
- State the definition of a choice function for a family \((X_i)_{i\in I}\), and prove from the definition that the product is nonempty if and only if such a function exists.
- Why does the compactness of every factor not imply that the product has a point? Give a concrete example, including the topology on each factor.
- State the axiom of choice in terms of products of nonempty sets. Does Tychonoff's compactness conclusion itself assert that such a product is nonempty?