Proving Connectedness by Ruling Out a Separation
A direct proof that a space is connected must rule out every possible separation, not merely one convenient partition. This can make the definition feel difficult to use. A contradiction proof reverses the task: assume that a separation exists, follow how its pieces intersect natural subsets of the space, and show that the assumed split cannot persist.
Cartesian products provide a useful setting for this strategy. Their horizontal and vertical slices inherit connectedness from the factors. These slices overlap in a way that forces any proposed separation of the whole product to place every point on the same side. The contradiction is that a separation must have two nonempty sides.
Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces. We use the maximum metric on their Cartesian product:
A ball for this metric restricts both coordinates at once. In particular, the maps that keep one coordinate fixed are continuous, so a connected factor gives connected slices in the product. We will use the theorem on continuous images of connected sets and the theorem that a connected subset lies on one side of a separation in a metric space, established earlier in this course.
Proof. Suppose, for a contradiction, that \(X\times Y\) is disconnected. Then it has a separation \(X\times Y=A\cup B\), with \(A\) and \(B\) nonempty and separated. Choose \(x_0\in X\) and \(y_0\in Y\). The point \((x_0,y_0)\) belongs to one of the pieces; after interchanging their names if necessary, suppose \((x_0,y_0)\in A\).
First consider the vertical slice \(\{x_0\}\times Y\). The map \(y\mapsto(x_0,y)\) from \(Y\) into \(X\times Y\) is continuous: for any \(y,y'\in Y\), \(d((x_0,y),(x_0,y'))=d_Y(y,y')\). Its image, the slice, is therefore connected. It meets \(A\) at \((x_0,y_0)\). By the theorem that a connected subset lies on one side of a separation, the whole slice must lie in \(A\). Thus \((x_0,y)\in A\) for every \(y\in Y\).
Now fix any \(y\in Y\) and consider the horizontal slice \(X\times\{y\}\). The map \(x\mapsto(x,y)\) is continuous, since \(d((x,y),(x',y))=d_X(x,x')\). Because \(X\) is connected, this slice is connected. It meets \(A\) at \((x_0,y)\), so again it lies entirely in \(A\). Since \(y\) was arbitrary, every point of \(X\times Y\) belongs to \(A\).
This gives \(B=\varnothing\), contradicting the requirement that both pieces of a separation be nonempty. Hence \(X\times Y\) is connected. \(\square\)
Why the Contradiction Works
The proof uses more than the fact that each factor is connected. The slices must be linked: every horizontal slice meets the vertical slice through \(x_0\), and that vertical slice meets the chosen point \((x_0,y_0)\). A connected slice cannot cross a separation, so the choice of side made at the anchor point propagates first along the vertical slice and then across every horizontal slice.
Write the product as two nonempty separated pieces and choose an anchor point in one piece.
The vertical slice through the anchor is connected and meets the chosen piece, so it lies entirely in that piece.
Every horizontal slice meets the vertical slice, so each lies in the same piece.
The supposed other piece has no points, contrary to the definition of a separation.
This argument depends on the factors being nonempty. If either factor is empty, the product is empty and has no separation, even if the other factor is disconnected. Thus the product is connected whenever both factors are nonempty and connected, but the nonempty hypothesis cannot be omitted from a converse about the factors.
Worked Example: A Closed Rectangle Is Connected
Consider \(R=[-2,1]\times[3,5]\), with the maximum metric inherited from \(\mathbb{R}^2\). Both coordinate intervals are nonempty and connected by the theorem that intervals are connected. The product theorem therefore gives that \(R\) is connected.
To see how the contradiction argument applies here, suppose \(R=A\cup B\) were a separation, and choose an anchor \((x_0,y_0)\in A\). The vertical slice \(\{x_0\}\times[3,5]\) is connected, so it must lie in \(A\). Each horizontal slice \([-2,1]\times\{y\}\) is connected and meets that vertical slice at \((x_0,y)\), so it too must lie in \(A\). These horizontal slices cover \(R\), forcing \(B\) to be empty. This contradicts the supposed separation.
Worked Example: A Product with an Unbounded Factor
The intervals \([0,2]\) and \((-\infty,0)\) are nonempty and connected. The product theorem shows that \([0,2]\times(-\infty,0)\) is connected with the maximum metric. No boundedness or compactness assumption is needed: the proof depends on connected slices and their intersections, not on the size of the factors.
In particular, take any proposed separation of this product and choose \((x_0,y_0)\) in its first piece. The vertical slice \(\{x_0\}\times(-\infty,0)\) remains connected even though it is unbounded, and therefore lies in that piece. Every horizontal slice \([0,2]\times\{y\}\) then meets it and must lie there as well. Thus the other piece would be empty. The same argument works for unbounded factors because connectedness of a slice does not require the slice to be bounded.
The Converse and Finite Products
For nonempty factors, connectedness of the product also forces connectedness of each factor. The coordinate projections \(\pi_X(x,y)=x\) and \(\pi_Y(x,y)=y\) are continuous for the maximum metric: for example, \(d_X(\pi_X(x,y),\pi_X(x',y'))=d_X(x,x')\leq d((x,y),(x',y'))\). If both factors are nonempty, each projection maps \(X\times Y\) onto its factor. The continuous-image theorem then implies that if \(X\times Y\) is connected, both \(X\) and \(Y\) are connected.
Proof. If both factors are connected, the Connectedness of a Product of Two Metric Spaces applies. Conversely, suppose \(X\times Y\) is connected. Nonemptiness of both factors makes each coordinate projection onto its factor. The projections are continuous, and continuous images of connected sets are connected. Hence \(X\) and \(Y\) are connected. \(\square\)
Repeated application gives a finite-product result. Equip \(X_1\times\cdots\times X_n\) with the metric \(d((x_i),(x_i'))=\max_{1\leq i\leq n}d_i(x_i,x_i')\), where \(d_i\) is the metric on \(X_i\).
Proof. We use induction on the positive integer \(n\). For \(n=1\), the assertion is exactly the connectedness of \(X_1\). Suppose the assertion holds for \(n\) factors. The product \(P=X_1\times\cdots\times X_n\) is nonempty and connected by the induction hypothesis. The two-factor theorem applied to \(P\) and \(X_{n+1}\) shows that \(P\times X_{n+1}\) is connected. Its maximum metric agrees with the stated maximum metric on \(X_1\times\cdots\times X_{n+1}\), since taking the maximum first over the first \(n\) coordinate distances and then with the last distance gives the maximum over all \(n+1\) distances. This completes the induction. \(\square\)
Worked Example: A Three-Dimensional Box
Let \(Q=[0,1]\times[2,4]\times[-3,0]\), equipped with the maximum metric on \(\mathbb{R}^3\). Each factor is a nonempty interval and hence connected. The finite-product theorem, with three factors, shows that \(Q\) is connected.
The induction proof describes the reasoning: first, \([0,1]\times[2,4]\) is connected by the two-factor theorem. Then its product with \([-3,0]\) is connected by another application of the same theorem. The conclusion does not require a separate argument about points in the box or a direct search for separating sets.
When a Product Is Disconnected
The nonempty product criterion also helps diagnose failure. If one nonempty factor is disconnected, the product cannot be connected. It is useful to see an explicit separation as well: it confirms that the obstruction is a genuine split in the product, rather than merely a failure of a theorem’s hypothesis.
Worked Example: Two Horizontal Pieces Do Not Form a Connected Product
Let \(E=[0,1]\times\{-1,2\}\), with the maximum metric. Define \(A=[0,1]\times\{-1\}\) and \(B=[0,1]\times\{2\}\). The two sets are nonempty, disjoint, and cover \(E\). For any \(a=(x,-1)\in A\) and \(b=(x',2)\in B\), the distance is \(d(a,b)=\max\{|x-x'|,3\}\geq3\). Thus \(\operatorname{dist}(A,B)\geq3>0\), so the positive-distance separation theorem shows that \(A\) and \(B\) are separated. Consequently \(E\) is disconnected.
This example is consistent with the product criterion: the factor \(\{-1,2\}\) is disconnected, and both factors are nonempty. The gap of \(3\) between the second coordinates makes the separation visible directly. The theorem does not claim that every product is connected; it identifies connectedness of both nonempty factors as the precise condition.
A Proof-Checking Habit
In contradiction proofs involving connectedness, the assumed separation has several requirements: its two pieces cover the space, are disjoint, are separated (or relatively open), and are nonempty. The product proof uses the last requirement only at the final step, when it forces the purported second piece to be empty. Forgetting that final check would leave the contradiction incomplete.
The central technique is to find connected subsets that overlap in a controlled way. The theorem that a connected subset lies on one side of a separation then transfers the chosen side from one subset to another. In products, slices provide exactly this structure. The method is useful beyond rectangles: whenever connected subsets cover a space and their intersections link them to a common anchor, a supposed separation may be forced to place the whole space on one side.
Check Your Understanding
Use the product theorem and its proof strategy to answer the following questions.
- Why does the connectedness proof begin by assuming a separation of the product?
- Which continuous maps show that the vertical and horizontal slices are connected?
- Why does each horizontal slice have to lie in the same piece as the anchor point?
- Where does the proof use the assumption that both factors are nonempty?
- How does the coordinate-projection argument prove the converse for nonempty factors?
- For the product \([0,1]\times\{0,4\}\), identify two nonempty pieces that form a separation.