The Coarsest Topology Making Projections Continuous

Why the Projections Determine a Topology

A Cartesian product has many coordinate projections: each one reads a single coordinate of a point. Requiring those maps to be continuous imposes a condition on which subsets of the product can be open. The product topology is precisely the least topology satisfying all of those conditions. The key is to collect the required open sets as a subbasis, then check both that the resulting topology works and that every competing topology must contain it.

Let \(J\) be an index set, let \((X_\alpha,\mathcal{T}_\alpha)\) be a topological space for each \(\alpha\in J\), and let \(X=\prod_{\alpha\in J}X_\alpha\) be their Cartesian product as a set. For each \(\alpha\), the coordinate projection is the map \(\pi_\alpha:X\to X_\alpha\) defined by \(\pi_\alpha((x_\beta)_{\beta\in J})=x_\alpha\). If some factor is empty, then \(X\) is empty; the arguments below still apply.

Definition: The projection-generated subbasis on \(X\) consists of all sets \(\pi_\alpha^{-1}(U)\), where \(\alpha\in J\) and \(U\in\mathcal{T}_\alpha\). The topology generated by this subbasis is the collection of arbitrary unions of finite intersections of these sets. The empty intersection is \(X\), and the empty union is \(\varnothing\). This generated topology is the product topology on \(X\).

A finite intersection of subbasic sets restricts only finitely many coordinates. If the same coordinate occurs more than once, its restrictions combine by intersection. Thus a typical basic open set has the form \(\bigcap_{k=1}^{n}\pi_{\alpha_k}^{-1}(U_k)\), where each \(U_k\) is open in its corresponding factor. Coordinates not listed among the \(\alpha_k\) are unrestricted. When \(J\) is empty, there are no subbasic sets; the product is a singleton and the generated topology is \(\{\varnothing,X\}\).

The Coarsest-Topology Characterization

Theorem: The product topology on \(X=\prod_{\alpha\in J}X_\alpha\) is the coarsest topology on \(X\) for which every coordinate projection \(\pi_\alpha:X\to X_\alpha\) is continuous. Here “coarsest” means that it is contained in every topology on \(X\) with this property.

Proof. Let \(\mathcal{T}_{\mathrm{prod}}\) denote the topology generated by the projection-generated subbasis. First, fix \(\alpha\in J\) and let \(U\in\mathcal{T}_\alpha\). By construction, \(\pi_\alpha^{-1}(U)\in\mathcal{T}_{\mathrm{prod}}\). Continuity means that the preimage of every open set is open, so \(\pi_\alpha\) is continuous when \(X\) has the product topology. This holds for each \(\alpha\).

Now let \(\mathcal{T}'\) be any topology on the set \(X\) for which every \(\pi_\alpha\) is continuous. For each \(\alpha\) and each \(U\in\mathcal{T}_\alpha\), continuity gives \(\pi_\alpha^{-1}(U)\in\mathcal{T}'\). Hence \(\mathcal{T}'\) contains every member of the projection-generated subbasis. Since \(\mathcal{T}'\) is a topology, it contains every finite intersection of those members and every union of such intersections. It therefore contains \(\mathcal{T}_{\mathrm{prod}}\). If \(J\) is empty, the first assertion has no projections to check, and the only topology on the singleton product is \(\{\varnothing,X\}\), so the same conclusion holds. Thus the product topology makes all projections continuous and is contained in every topology that does so. \(\square\)

The proof has two distinct directions. The subbasis construction guarantees that the projections are continuous. Conversely, continuity forces every subbasic preimage to be open, and the topology axioms then force all product-open sets to be open. Neither direction alone establishes the characterization.

Worked Example: Open Rectangles in \(\mathbb{R}^2\)

Give both factors the standard topology on \(\mathbb{R}\). The product set is \(\mathbb{R}\times\mathbb{R}\), and its projections are \(\pi_1(x,y)=x\) and \(\pi_2(x,y)=y\). For open intervals \(U=(-2,1)\) and \(V=(3,5)\), compute the subbasic preimages and their intersection.

By the definition of a preimage, \(\pi_1^{-1}(U)\) consists of all pairs whose first coordinate lies in \(U\), so \(\pi_1^{-1}(U)=(-2,1)\times\mathbb{R}\). Likewise, \(\pi_2^{-1}(V)=\mathbb{R}\times(3,5)\). Intersecting imposes both restrictions at once:

\pi_1^{-1}((-2,1))\cap\pi_2^{-1}((3,5)) =((-2,1)\times\mathbb{R})\cap(\mathbb{R}\times(3,5)) =(-2,1)\times(3,5).

Check: a pair \((x,y)\) belongs to the intersection exactly when \(-2<x<1\) and \(3<y<5\), which is exactly the membership condition for the rectangle on the right. The rectangle is product-open because it is a finite intersection of subbasic open sets. The theorem also explains why any topology making both projections continuous must include this rectangle as an open set.

Finite Intersections and Arbitrary Products

The finite-intersection condition matters even when there are infinitely many factors. A subbasic open set places a condition on one coordinate. A basic open set can combine finitely many such conditions, but it does not impose conditions on infinitely many coordinates at once. This follows directly from how a topology is generated by a subbasis: the required building blocks are finite intersections, followed by arbitrary unions.

Worked Example: Restrictions on Two Coordinates of a Sequence

Let \(X_\alpha=\mathbb{R}\) with its standard topology for every \(\alpha\in\mathbb{N}\). A point of \(X=\prod_{n\in\mathbb{N}}\mathbb{R}\) is a sequence \(x=(x_1,x_2,\dots)\). Consider the set \(W\) of sequences satisfying \(0<x_2<4\) and \(-1<x_5<2\), with no restrictions on other coordinates.

The restriction on the second coordinate is \(\pi_2^{-1}((0,4))\), and the restriction on the fifth coordinate is \(\pi_5^{-1}((-1,2))\). Thus

W=\pi_2^{-1}((0,4))\cap\pi_5^{-1}((-1,2)).

This is a finite intersection of subbasic open sets, so it is open in the product topology. For example, the sequence \(x=(0,1,2,3,0,6,\dots)\) belongs to \(W\): its second coordinate is \(1\), with \(0<1<4\), and its fifth coordinate is \(0\), with \(-1<0<2\). A sequence whose second coordinate is \(4\) does not belong to \(W\), because the interval \((0,4)\) excludes its endpoint \(4\); a sequence whose fifth coordinate is \(-1\) also does not belong, because \((-1,2)\) excludes \(-1\).

Check: membership in \(W\) is equivalent to the two stated coordinate conditions, regardless of every other coordinate. The set is therefore exactly the intersection displayed above, not a set that restricts all coordinates.

A Small Product with Its Generated Topology

Worked Example: A Product of Two Finite Spaces

Let \(X_1=\{a,b\}\) have topology \(\{\varnothing,\{a\},X_1\}\), and let \(X_2=\{0,1\}\) have the discrete topology \(\mathcal{P}(X_2)\). Find the subbasic sets and describe the topology they generate on \(X_1\times X_2\).

Write \(A=\{a\}\times X_2\), \(C_0=X_1\times\{0\}\), and \(C_1=X_1\times\{1\}\). The nonempty proper open set \(\{a\}\) in the first factor gives the subbasic set \(A\). The singleton opens in the discrete second factor give \(C_0\) and \(C_1\). The whole-space opens in either factor give \(X_1\times X_2\), and empty opens give \(\varnothing\). So the subbasis is \(\{A,C_0,C_1\}\), together with \(\varnothing\) and the whole product if those are included explicitly.

Finite intersections produce \(\{(a,0)\}=A\cap C_0\) and \(\{(a,1)\}=A\cap C_1\), as well as \(A,C_0,C_1\), and the whole product. Taking unions gives exactly the following open sets:

\varnothing,\ \{(a,0)\},\ \{(a,1)\},\ A,\ C_0,\ C_1,\ A\cup C_0,\ A\cup C_1,\ X_1\times X_2.

To check that the list is complete, every finite intersection of \(A,C_0,C_1\) is one of the listed sets: \(C_0\cap C_1=\varnothing\), while \(A\cap C_0=\{(a,0)\}\) and \(A\cap C_1=\{(a,1)\}\). Unions of the listed generators yield only the sets displayed. In particular, \(\{(b,0)\}=C_0\setminus A\) is not open in this generated topology: it cannot be obtained by unions of finite intersections, and every listed open set containing \((b,0)\) also contains \((a,0)\).

Check: the preimage of \(\{a\}\) under the first projection is \(A\), which is open. For the second projection, the preimages of \(\{0\}\) and \(\{1\}\) are \(C_0\) and \(C_1\), both open. The preimages of the empty set and the whole factor are also open. Thus both projections are continuous, as the theorem predicts.

Common Pitfall: Reversing the Inclusion

“Coarsest” can make the direction of the inclusion easy to reverse. If \(\mathcal{T}'\) makes every projection continuous, then it must contain the product topology: \(\mathcal{T}_{\mathrm{prod}}\subseteq\mathcal{T}'\). Continuity requires preimages of factor-open sets to be open, so it imposes required open sets rather than removing them.

For a concrete failure, give \(\mathbb{R}^2\) the indiscrete topology \(\{\varnothing,\mathbb{R}^2\}\) and give each copy of \(\mathbb{R}\) its standard topology. The set \((0,1)\) is open in the first factor, but its preimage under \(\pi_1\) is \((0,1)\times\mathbb{R}\). This preimage is nonempty and is not all of \(\mathbb{R}^2\), so it is not open in the indiscrete topology. Therefore \(\pi_1\) is not continuous for that topology, and the indiscrete topology cannot serve as a topology making the projections continuous.

At the other extreme, the discrete topology on \(\mathbb{R}^2\) makes every map out of \(\mathbb{R}^2\) continuous, since every subset of the domain is open. It therefore makes the projections continuous, but the theorem says only that the product topology is contained in it, not that the two topologies agree. The characterization identifies a least topology, not the only qualifying topology.

Takeaway: To identify the coarsest topology making maps continuous, collect the preimages of open sets in their codomains. The topology generated by those preimages is the least possible one: it makes the maps continuous, and every topology that does so must contain it.

Check Your Understanding

Use the projection-generated subbasis and the coarsest-topology theorem to answer these questions.

For \(X=\mathbb{R}\times\mathbb{R}\) with standard factor topologies, write \((1,3)\times(-2,0)\) as an intersection of two subbasic open sets.
In a product indexed by \(\mathbb{N}\), explain why a restriction on the first four coordinates gives a basic open set when each restriction is open in its factor.
Suppose \(\mathcal{T}'\) makes every coordinate projection continuous. Which topology is contained in \(\mathcal{T}'\), and why?
In the finite product example, why is \(\{(b,0)\}\) not open in the generated topology?
What topology is generated on the product of an empty family of spaces, and why?