Why Finite Coordinates Control the Product
In a product with many coordinates, a basic open set in the product topology restricts only finitely many coordinates at a time. This feature makes it possible to build a dense subset by allowing only finitely many coordinates to vary. When every factor is connected, each finite-coordinate piece is connected; the pieces fit together at one common point. Their union is therefore connected, and its density transfers connectedness to the whole product.
Let \(I\) be an index set, and let \((X_i,\mathcal{T}_i)\) be a topological space for each \(i\in I\). The set underlying the product is \(X=\prod_{i\in I}X_i\), and \(\pi_i:X\to X_i\) denotes the \(i\)-th coordinate projection. The product topology has as a basis the sets \(\bigcap_{i\in E}\pi_i^{-1}(U_i)\), where \(E\subseteq I\) is finite and each \(U_i\) is open in \(X_i\). For \(E=\varnothing\), this basis set is all of \(X\).
The product with no factors is, by convention, a one-point space. For the theorem below, assume each factor is nonempty and connected. If a product of such factors is empty, it is connected by the definition. If it is nonempty, it has a point \(a=(a_i)_{i\in I}\); fix that point. In the usual set theory with the axiom of choice, a product of nonempty sets is nonempty. The proof here only needs to choose \(a\) in the product when the product is nonempty.
The Finite-Coordinate Construction
For each finite subset \(F\subseteq I\), let \(X_F\) consist of the points of \(X\) whose coordinates outside \(F\) agree with \(a\). Thus coordinates in \(F\) may vary, while all other coordinates are fixed.
As a subspace of \(X\), \(X_F\) has the same topology as the finite product \(\prod_{i\in F}X_i\). Indeed, inserting the fixed coordinates \(a_i\) gives a bijection from that finite product to \(X_F\). A basic open set in \(X\), when intersected with \(X_F\), either misses \(X_F\) because a restriction outside \(F\) excludes its fixed value \(a_i\), or restricts only finitely many coordinates in \(F\). Conversely, any basic open set of the finite product is obtained by imposing those same restrictions in \(X\). This verifies the agreement of the subspace and finite-product topologies.
Proof. If \(F=\varnothing\), then \(X_F=\{a\}\), which is connected. If \(F\) has one element, \(X_F\) is homeomorphic to its single factor, which is connected by hypothesis. For any finite \(F\) with at least two elements, list its elements as \(i_1,\dots,i_n\). The result on products of two connected spaces gives that \(X_{i_1}\times X_{i_2}\) is connected. Applying that result successively, the product of the first \(k\) factors is connected for each \(k\) from \(2\) through \(n\): at each step it is the product of a connected space with the connected factor \(X_{i_k}\). Hence \(\prod_{i\in F}X_i\), and therefore its homeomorphic copy \(X_F\), is connected. \(\square\)
Now take the union of all these finite-coordinate subspaces:
Every \(X_F\) contains \(a\), so the subspaces have a common point. Their union is connected. To see this directly, suppose \(Y\) had a separation \(Y=U\cup V\), where \(U,V\) are disjoint, nonempty, and open in \(Y\). The point \(a\) lies in one of the two sets; suppose \(a\in U\). Each connected \(X_F\) must lie wholly in \(U\): otherwise \(X_F\cap U\) and \(X_F\cap V\) would be a separation of \(X_F\), since both intersections would be nonempty and relatively open. Thus every \(X_F\subseteq U\), so \(Y\subseteq U\), contradicting \(V\ne\varnothing\). Hence \(Y\) is connected.
Proof. Let \(W\) be a nonempty open subset of \(X\). It contains a nonempty basic open set \(B=\bigcap_{i\in E}\pi_i^{-1}(U_i)\), with \(E\) finite. Choose \(x\in B\), and define \(y\in X\) by setting \(y_i=x_i\) for \(i\in E\) and \(y_i=a_i\) for \(i\notin E\). Then \(y\in X_E\subseteq Y\). For each \(i\in E\), \(y_i=x_i\in U_i\), so \(y\in B\subseteq W\). Thus \(W\cap Y\ne\varnothing\), proving density. \(\square\)
Proof. If \(X=\prod_{i\in I}X_i\) is empty, it is connected by the definition. Otherwise, the construction above gives a connected dense subset \(Y\subseteq X\). It remains to verify that a space containing a dense connected subset is connected. If \(X\) had a separation \(X=U\cup V\), with \(U,V\) disjoint, nonempty, and open in \(X\), density would imply that both \(U\) and \(V\) meet \(Y\). Their intersections with \(Y\) would then be disjoint, nonempty, relatively open subsets covering \(Y\), a separation of \(Y\). This contradicts connectedness of \(Y\). Therefore \(X\) is connected. If \(I=\varnothing\), the product is a one-point space and is connected as stated. \(\square\)
Worked Products
Worked Example: Countably Many Real Lines
Consider \(X=\mathbb{R}^{\mathbb{N}}\), with each copy of \(\mathbb{R}\) carrying its standard topology and \(X\) carrying the product topology. Each factor is connected, since \(\mathbb{R}\) is an interval. Choose the zero sequence \(a_n=0\) for all \(n\). The finite-coordinate union \(Y\) consists exactly of sequences with only finitely many nonzero entries.
To check density concretely, take a basic open set that restricts coordinates \(2\) and \(5\) to \(U_2=(1,2)\) and \(U_5=(-1,0)\), respectively, and leaves every other coordinate unrestricted. Set \(y_2=3/2\), \(y_5=-1/2\), and \(y_n=0\) for all \(n\notin\{2,5\}\). The calculations \(1<3/2<2\) and \(-1<-1/2<0\) show that \(y\) belongs to the basic open set. Its only possibly nonzero entries are in \(\{2,5\}\), so \(y\in Y\). The same construction works for any finite collection of coordinate restrictions: choose a point in each restricted open set and put zero in every other coordinate. Thus every nonempty basic open set meets \(Y\).
The theorem gives that \(\mathbb{R}^{\mathbb{N}}\) is connected. As a check on the distinction between \(Y\) and \(X\), the sequence \((1,1,1,\dots)\) is in \(X\) but not in \(Y\), because it has infinitely many nonzero coordinates. The proof needs \(Y\) to be dense, not equal to \(X\).
Worked Example: An Uncountable Product of Intervals
Let \(J=\mathbb{R}\), and let \(X=\prod_{t\in J}[0,1]\), with each interval carrying its usual subspace topology and \(X\) the product topology. Each factor is connected because it is an interval. Choose the point \(a\) whose every coordinate is \(0\). The associated dense subset consists of functions \(J\to[0,1]\) that are zero except at finitely many indices.
For a specific basic open set, restrict coordinate \(1\) to \((1/4,3/4)\cap[0,1]\) and coordinate \(\pi\) to \((0,1/3)\cap[0,1]\), leaving all other coordinates unrestricted. Define \(y_1=1/2\), \(y_\pi=1/4\), and \(y_t=0\) for every \(t\notin\{1,\pi\}\). The first restricted coordinate satisfies \(1/4<1/2<3/4\); the second satisfies \(0<1/4<1/3\). Hence \(y\) belongs to the basic open set, and it is nonzero at only two indices. For any nonempty basic open set, its finite set of restrictions can be met by choosing one allowed value in each restricted coordinate and using zero elsewhere. The theorem therefore proves that this uncountable product is connected.
Check: both chosen values are in \([0,1]\), and the unrestricted coordinates have value \(0\in[0,1]\). The construction uses only finitely many nonzero coordinates, regardless of the size of \(J\).
Worked Example: A Product of Sierpinski Spaces
Let each factor be \(S=\{0,1\}\) with topology \(\{\varnothing,\{0\},S\}\), and consider \(S^{\mathbb{N}}\) in the product topology. The factor \(S\) is connected: every nonempty open set contains \(0\), so two disjoint nonempty open sets cannot exist. Choose the point \(a=(1,1,1,\dots)\). The finite-coordinate union consists of sequences equal to \(1\) except at finitely many coordinates.
A basic open set in this product restricts finitely many coordinates to open subsets of \(S\). If the basic open set is nonempty, each restriction is either \(S\) or \(\{0\}\); a restriction to the empty set would make it empty. Set each coordinate restricted to \(\{0\}\) equal to \(0\), and set every other coordinate equal to \(1\). This sequence satisfies every restriction. It differs from \(a\) at only the finitely many coordinates restricted to \(\{0\}\), so it belongs to the finite-coordinate union. Thus that union is dense, and the theorem gives that \(S^{\mathbb{N}}\) is connected.
Check: if no coordinate is restricted to \(\{0\}\), the constructed point is \(a\) itself and still belongs to the basic open set. If there are \(m\) such restrictions, the constructed sequence differs from \(a\) in exactly those \(m\) coordinates, which is finite.
A Common Pitfall: Dense Does Not Mean Equal
A mistaken argument might say that the union of finite-coordinate subproducts is the whole product. In \(\mathbb{R}^{\mathbb{N}}\) with basepoint the zero sequence, that would claim \((1,1,1,\dots)\in Y\). But this sequence differs from the basepoint in every coordinate, so there is no finite \(F\subseteq\mathbb{N}\) for which it belongs to \(X_F\). The claim is false. The correction is that \(Y\) is dense, as the basis-set calculation proves, and a dense connected subset forces the whole space to be connected.
The finite-coordinate condition is specific to the product topology: its basic open sets restrict only finitely many coordinates. In the box topology, basic open sets may restrict infinitely many coordinates, so the density argument above does not establish density there. The theorem proved here concerns the product topology, not the box topology.
Check Your Understanding
Use the finite-coordinate construction and the product-topology basis in your reasoning.
- For a product of connected spaces with a chosen point \(a\), describe \(X_F\) when \(F=\varnothing\).
- Why does a nonempty basic open set in the product topology have a point in the finite-coordinate union?
- In \(\mathbb{R}^{\mathbb{N}}\) with zero basepoint, explain why the sequence \((1,1,1,\dots)\) is not in the finite-coordinate union but does not obstruct density.
- Which step of the density proof would need to change for the box topology?
- Explain why a separation of the whole product would induce a separation of its dense connected subset.