Why Slices Control a Product
A product has two coordinate directions. Fixing one coordinate gives a slice that varies only in the other direction. If both spaces are connected, these slices are connected; moreover, every vertical slice meets a chosen horizontal slice. That pattern of intersections is enough to rule out a separation of the whole product.
Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces. The product topology on \(X\times Y\) has as a basis all rectangles \(U\times V\), where \(U\in\mathcal{T}_X\) and \(V\in\mathcal{T}_Y\). In particular, a set is open in \(X\times Y\) if it is a union of such rectangles.
We use the connected-image result: if \(C\) is connected and \(f:C\to Z\) is continuous, then \(f(C)\), with its subspace topology, is connected. Indeed, a separation of \(f(C)\) would have disjoint, nonempty, relatively open preimages under \(f\); because \(f\) maps onto \(f(C)\), those preimages would cover \(C\), giving a separation of \(C\). This is the standard continuous-image property of connectedness.
Connected Slices and Their Intersections
For \(y\in Y\), the horizontal slice at \(y\) is \(X\times\{y\}\). For \(x\in X\), the vertical slice at \(x\) is \(\{x\}\times Y\). When the factors are connected, each slice is connected as a subspace of the product.
Proof. Fix \(y\in Y\) and define \(i_y:X\to X\times Y\) by \(i_y(x)=(x,y)\). For a basic open rectangle \(U\times V\), its preimage is \(U\) if \(y\in V\), and is \(\varnothing\) if \(y\notin V\). Both are open in \(X\), so \(i_y\) is continuous: preimages of arbitrary open sets are open because every open set is a union of basic rectangles. Its image is \(X\times\{y\}\). The connected-image result therefore makes this slice connected. For a fixed \(x\in X\), define \(j_x:Y\to X\times Y\) by \(j_x(y)=(x,y)\). The preimage of \(U\times V\) is \(V\) if \(x\in U\), and \(\varnothing\) otherwise. Thus \(j_x\) is continuous, its image is \(\{x\}\times Y\), and that slice is connected as well. \(\square\)
Proof. If either factor is empty, then \(X\times Y\) is empty and hence connected. Now suppose both factors are nonempty. Choose \(y_0\in Y\), and suppose for contradiction that \(X\times Y\) has a separation \(P,Q\). The horizontal slice \(H=X\times\{y_0\}\) is connected by the lemma. Its intersections with \(P\) and \(Q\) are relatively open in \(H\), disjoint, and together cover \(H\). Connectedness of \(H\) means at least one intersection is empty. Since \(H\) is nonempty, it lies entirely in one side; after interchanging the names of the sides if necessary, assume \(H\subseteq P\).
For each \(x\in X\), the vertical slice \(V_x=\{x\}\times Y\) is connected. It meets \(P\), because \((x,y_0)\in V_x\cap H\subseteq P\). If \(V_x\) also met \(Q\), then \(V_x\cap P\) and \(V_x\cap Q\) would be disjoint, nonempty, relatively open subsets covering \(V_x\), contradicting its connectedness. Hence \(V_x\subseteq P\) for every \(x\in X\). Every \((x,y)\in X\times Y\) belongs to \(V_x\), so all of \(X\times Y\) lies in \(P\). This contradicts the requirement that \(Q\) be nonempty. No separation exists, and the product is connected. \(\square\)
The point of the proof is not that all vertical slices meet one another. They need not. Instead, each vertical slice meets the same connected horizontal slice, which forces every vertical slice to lie on the same side of any proposed separation.
Products of Intervals
Every interval in \(\mathbb{R}\) with the standard topology is connected. Thus the theorem applies to products of intervals, using the product topology. The first example spells out how the intersecting slices cover a rectangle, including its boundary.
Worked Example: A Closed Rectangle
Give the connectedness argument for \([0,2]\times[1,3]\), where each interval has the subspace topology inherited from the standard topology on \(\mathbb{R}\), and the product has the product topology.
The intervals \([0,2]\) and \([1,3]\) are connected, so the theorem says their product is connected. To see the slices in the proof explicitly, choose \(y_0=1\). The horizontal slice is \([0,2]\times\{1\}\). For each \(x\in[0,2]\), the vertical slice is \(\{x\}\times[1,3]\), and it meets the horizontal slice at exactly \((x,1)\): its first coordinate must be \(x\), and its second coordinate must be \(1\).
These vertical slices cover the product. In fact, for any \((s,t)\in[0,2]\times[1,3]\), we have \(s\in[0,2]\) and \(t\in[1,3]\), so \((s,t)\in\{s\}\times[1,3]\). The endpoints are included: for instance, \((0,1)\in\{0\}\times[1,3]\), and \((2,3)\in\{2\}\times[1,3]\). Thus every vertical slice meets the connected horizontal slice and together they cover the entire rectangle, exactly as required by the separation proof.
Check: the chosen level \(y_0=1\) belongs to \([1,3]\); the intersection point \((x,1)\) belongs to both slices for every \(x\in[0,2]\); and every point \((s,t)\) of the product belongs to the slice indexed by its first coordinate \(s\).
Worked Example: An Unbounded Product
Show that \(\mathbb{R}\times[0,1]\) is connected, taking \(\mathbb{R}\) with its standard topology, \([0,1]\) with its subspace topology, and their product with the product topology.
The real line is an interval, and \([0,1]\) is an interval, so both factors are connected. Choose \(y_0=0\). The horizontal slice \(\mathbb{R}\times\{0\}\) is connected. For each \(x\in\mathbb{R}\), the vertical slice \(\{x\}\times[0,1]\) is connected and meets the horizontal slice at \((x,0)\). If \((s,t)\) is any point of the product, then \(s\in\mathbb{R}\) and \(t\in[0,1]\), so \((s,t)\in\{s\}\times[0,1]\). The same separation argument therefore proves connectedness, with no boundedness assumption on the first factor.
Check: \(0\in[0,1]\), so each intersection point \((x,0)\) is in both slices. The value \(t\) of any point in the product satisfies \(0\leq t\leq1\), so it lies in the indicated vertical slice.
Worked Example: A Degenerate Interval Factor
Consider \([2,2]\times[0,1]\), with the subspace topologies from the standard topology on \(\mathbb{R}\) and the product topology. This product is connected.
The interval \([2,2]\) is the singleton \(\{2\}\), which is connected: a singleton cannot be split into two nonempty disjoint subsets. The interval \([0,1]\) is connected. Choose \(y_0=0\). The horizontal slice \(\{2\}\times\{0\}\) meets the only vertical slice, \(\{2\}\times[0,1]\), at \((2,0)\). The vertical slice is the whole product, so it covers every point \((2,t)\) with \(t\in[0,1]\). The theorem applies even when one factor has just one point.
Check: the only possible first coordinate is \(2\), and the only horizontal level chosen is \(0\in[0,1]\); hence the intersection point belongs to both slices. No nonempty factor or nondegeneracy assumption is needed beyond those already handled in the theorem.
Why Both Factors Matter
For nonempty factors, the theorem has a converse: if \(X\times Y\) is connected, then each factor is connected. To prove this, the coordinate projections \(\pi_X:X\times Y\to X\) and \(\pi_Y:X\times Y\to Y\), defined by \(\pi_X(x,y)=x\) and \(\pi_Y(x,y)=y\), are continuous. For example, \(\pi_X^{-1}(U)=U\times Y\), which is open in the product topology for every open \(U\subseteq X\). The corresponding preimage for \(\pi_Y\) is \(X\times V\), also open. If both factors are nonempty, these projections are onto.
Proof. If both factors are connected, the theorem proves that the product is connected. Conversely, suppose the product is connected and both factors are nonempty. Each projection is a continuous map from the product onto its factor. By the continuous-image property, the image of the connected product under each projection is connected. Those images are \(X\) and \(Y\), respectively. Thus both factors are connected. \(\square\)
Nonemptiness matters for this converse. If \(X=\varnothing\), then \(X\times Y=\varnothing\) is connected regardless of whether \(Y\) is connected; the projection onto \(Y\) is not onto when \(Y\) is nonempty. The forward theorem itself already covers this empty-product case.
Worked Example: A Disconnected Factor
Let \(X=\{0,1\}\) have the discrete topology (every subset is open), and let \(Y=[0,1]\) have its subspace topology from the standard topology on \(\mathbb{R}\). The product has the product topology.
The sets \(P=\{0\}\times[0,1]\) and \(Q=\{1\}\times[0,1]\) are disjoint, nonempty, and cover \(X\times Y\). They are open: \(P=\{0\}\times Y\) is a basic open rectangle because \(\{0\}\) is open in the discrete topology and \(Y\) is open in itself; likewise \(Q=\{1\}\times Y\) is open. Therefore \(P,Q\) form a separation, so the product is disconnected. This does not contradict the theorem: \(X\) is disconnected, since \(\{0\}\) and \(\{1\}\) themselves separate it.
Check: \(P\cap Q=\varnothing\) because no first coordinate is both \(0\) and \(1\); their union contains every point \((x,y)\) because \(x\in\{0,1\}\); and each contains points, for example \((0,0)\in P\) and \((1,0)\in Q\).
A Common Pitfall
Common pitfall. One might claim that all vertical slices have a common point, then use a common-point union argument. In \([0,2]\times[1,3]\), the vertical slices at \(0\) and \(2\) are \(\{0\}\times[1,3]\) and \(\{2\}\times[1,3]\). Their intersection is empty, since a point in the first has first coordinate \(0\), while a point in the second has first coordinate \(2\). So the proposed common-point claim is false.
The correction is to use one connected horizontal slice, not a common point shared by all vertical slices. Each vertical slice \(\{x\}\times[1,3]\) meets \([0,2]\times\{1\}\) at its own point \((x,1)\). In a separation, the horizontal slice must lie on one side; connectedness then forces each vertical slice, because it meets that side, to lie there too. This is the intersection mechanism that makes the product argument work.
Check Your Understanding
Use the slice argument and the definitions of connectedness and the product topology in your answers.
- For connected spaces \(X,Y\), choose \(x_0\in X\). Which horizontal or vertical slice can be used as the common connected slice, and where does each other slice meet it?
- Explain why a separation of \(X\times Y\) would restrict to a separation of any slice that meets both sides.
- Is \([1,4]\times\{7\}\) connected when both factors have their usual subspace topologies? Justify your answer, including the role of the singleton factor.
- Why does the converse proposition require both factors to be nonempty?
- In the discrete topology on \(\{a,b\}\), describe a separation of \(\{a,b\}\times[0,1]\) and verify that its two sets are open in the product topology.