Products of Connected Spaces

How Connectedness Behaves Under Products

A connected set cannot be split into two nonempty separated pieces. A product, however, has many directions in which one might try to move: one coordinate can vary while the others stay fixed. The key observation is that fixing one coordinate gives a copy of the other factor, so each such slice is connected when that factor is connected. We will show how these slices fit together to establish connectedness of the whole product.

Throughout, products of two metric spaces are equipped with the maximum metric. This choice makes the coordinate slices easy to describe and gives continuous coordinate projections. The main result is that the product of two connected metric spaces is connected. We will also prove a converse when both factors are nonempty and extend the result to any finite number of factors.

Definition: Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. Their product $X\times Y$ is equipped here with the maximum metric $$ d_{\max}((x,y),(x',y'))=\max\{d_X(x,x'),d_Y(y,y')\}. $$ For fixed $y_0\in Y$, the set $X\times\{y_0\}$ is called a horizontal slice; for fixed $x_0\in X$, the set $\{x_0\}\times Y$ is called a vertical slice.

A horizontal slice is the image of $X$ under the map $i_{y_0}(x)=(x,y_0)$. This map preserves distances, since $$ d_{\max}(i_{y_0}(x),i_{y_0}(x'))=\max\{d_X(x,x'),0\}=d_X(x,x'). $$ It is therefore continuous. Likewise, $y\mapsto(x_0,y)$ is continuous and maps $Y$ onto the vertical slice. By the Continuous Images of Connected Spaces Are Connected Theorem, each slice is connected whenever its varying factor is connected.

The Product of Two Connected Spaces

A single horizontal slice does not cover the product. To reach points off that slice, attach vertical slices. Each vertical slice meets the horizontal slice, and the union of two connected sets with a common point is connected by the Union with a Common Point Theorem. The resulting unions all contain the same horizontal slice, so they can be assembled into the entire product.

Theorem (Products of Two Connected Metric Spaces): If $X$ and $Y$ are nonempty connected metric spaces, then $X\times Y$, equipped with the maximum metric, is connected.

Proof. Choose $y_0\in Y$. The horizontal slice $H=X\times\{y_0\}$ is connected, because it is the continuous image of the connected space $X$. For each $x\in X$, the vertical slice $V_x=\{x\}\times Y$ is connected, because it is the continuous image of $Y$. The sets $H$ and $V_x$ have the common point $(x,y_0)$, so the Union with a Common Point Theorem shows that $C_x=H\cup V_x$ is connected.

Since $X$ is nonempty, choose $x_0\in X$. The point $(x_0,y_0)$ belongs to $H$, and hence belongs to every $C_x$. Thus the family $\{C_x:x\in X\}$ is a nonempty family of connected sets with a common point. A second application of the Union with a Common Point Theorem shows that $\bigcup_{x\in X}C_x$ is connected. Every point $(x,y)\in X\times Y$ belongs to $V_x\subseteq C_x$, and every $C_x$ lies in $X\times Y$. Consequently, $$ \bigcup_{x\in X}C_x=X\times Y. $$ Therefore $X\times Y$ is connected. $\square$

Worked Example: A Rectangle Is Connected as a Product

Consider $R=[-1,2]\times[0,3]$ with the maximum metric. Both $[-1,2]$ and $[0,3]$ are intervals in $\mathbb{R}$, so each is connected by the Theorem that Intervals in $\mathbb{R}$ Are Connected. They are also nonempty. The Products of Two Connected Metric Spaces Theorem therefore implies that $R$ is connected.

The conclusion follows from the connectedness of the two coordinate intervals and the way their slices overlap; it does not require us to construct a separation test for every possible pair of subsets of the rectangle. The maximum metric specifies the product metric but does not restrict the conclusion to that metric alone: any metric on this same product that gives the same topology has the same connected subsets.

Projections Give a Converse

Connectedness of the product also constrains its factors, provided neither factor is empty. Define the coordinate projections by $\pi_X(x,y)=x$ and $\pi_Y(x,y)=y$. For any two points of the product, $$ d_X(\pi_X(x,y),\pi_X(x',y'))=d_X(x,x')\leq d_{\max}((x,y),(x',y')), $$ and the corresponding inequality holds for $\pi_Y$. Thus both projections are continuous. If the factors are nonempty, each projection is onto, so each factor is a continuous image of the product.

Theorem (Connectedness of a Product and Its Factors): Let $X$ and $Y$ be nonempty metric spaces. Then $X\times Y$, with the maximum metric, is connected if and only if both $X$ and $Y$ are connected.

Proof. If both factors are connected, the Products of Two Connected Metric Spaces Theorem proves that $X\times Y$ is connected. Conversely, suppose $X\times Y$ is connected. The coordinate projections are continuous, and because both factors are nonempty, they are onto. The Continuous Images of Connected Spaces Are Connected Theorem now implies that $\pi_X[X\times Y]=X$ and $\pi_Y[X\times Y]=Y$ are connected. This proves both directions. $\square$

Worked Example: A Disconnected Factor Disconnects the Product

Give $\{-1,1\}$ the metric inherited from $\mathbb{R}$, and consider $P=\mathbb{R}\times\{-1,1\}$ with the maximum metric. Define $$ A=\mathbb{R}\times\{-1\},\qquad B=\mathbb{R}\times\{1\}. $$ These sets are disjoint, nonempty, and have union $P$. They are open in $P$: for any $(x,-1)\in A$, the ball in $P$ of radius $1$ around $(x,-1)$ contains no point with second coordinate $1$, since that coordinate differs by distance $2$. Hence this ball lies in $A$. The same argument shows that every point of $B$ has a radius-$1$ ball in $P$ contained in $B$.

Because $A$ and $B$ are complements in $P$, each is also relatively closed in $P$. They form a separation, so $P$ is disconnected. This is consistent with the theorem: the factor $\{-1,1\}$ is disconnected. The nonempty-factor hypothesis in the converse matters. If one factor is empty, the product is empty and is connected by convention, regardless of whether the other factor is connected.

Finite Products

The two-factor result can be applied repeatedly. For a finite list of metric spaces $X_1,\ldots,X_n$, use the maximum product metric $$ d((x_1,\ldots,x_n),(y_1,\ldots,y_n))=\max_{1\leq i\leq n}d_i(x_i,y_i). $$ Grouping the first $n-1$ coordinates together gives the same metric as taking the maximum of the metric on those coordinates and the distance in the last coordinate. This lets us apply the two-factor theorem at each stage.

Theorem (Finite Products of Connected Metric Spaces): A finite product of connected metric spaces, equipped with the maximum product metric, is connected. If any factor is empty, the product is empty and hence connected by convention.

Proof. If any factor is empty, the product is empty and the conclusion follows from the convention that the empty space is connected. Now suppose every factor is nonempty and connected. We argue by induction on $n$, the number of factors. For $n=1$, the statement is just the assumed connectedness of $X_1$. Suppose the result holds for $n-1$ factors. The product $X_1\times\cdots\times X_{n-1}$ is nonempty and connected by the induction hypothesis. Treat it as one metric space with its maximum metric, and apply the Products of Two Connected Metric Spaces Theorem to it and $X_n$. The resulting product is connected. Its metric is precisely the maximum of the $n$ coordinate distances, so this proves the claim for $n$. By induction the result holds for every finite $n$. $\square$

Worked Example: A Three-Dimensional Product of Intervals

Let $Q=[-1,1]\times(0,\infty)\times[2,4]$, using the maximum of the three usual coordinate distances. Each factor is a nonempty interval in $\mathbb{R}$, so each is connected. The Finite Products of Connected Metric Spaces Theorem shows that $Q$ is connected.

The open endpoint at $0$ in the second factor causes no difficulty: $(0,\infty)$ is still an interval, and intervals are connected whether or not their endpoints are included. The product theorem concerns connectedness, not compactness or closedness; neither boundedness nor inclusion of every endpoint is needed here.

Why the Slice Argument Is Useful

The proof of product connectedness is an example of a broader strategy: build a large set from connected pieces that overlap. A horizontal slice alone is connected, and each vertical slice is connected, but the overlap is what allows the pieces to be combined without creating a separation. It is not enough merely to know that a space is a union of connected subsets. For instance, the two sets $A$ and $B$ in the preceding example are each connected, but their disjoint union is not connected. The shared points in the slice argument are essential.

There are two hypotheses worth keeping distinct. For the forward implication, both factors must be connected. For the converse, both factors must be nonempty so that each projection reaches its whole factor. An empty product is connected by convention, but its projection cannot provide information about a nonempty factor. For finite products, the same caution applies: a product with an empty factor is empty, while for nonempty factors connectedness of the product is equivalent to connectedness of every factor.

Check Your Understanding

Use the slice argument and projection results to answer the following questions.

Why is the map $x\mapsto(x,y_0)$ continuous for the maximum metric?
In the proof of the two-factor theorem, why is each set $H\cup V_x$ connected?
Why do the connected sets $H\cup V_x$ have a common point as $x$ varies?
Which hypotheses ensure that the coordinate projections are onto?
Why does the converse fail to provide information about a factor when the other factor is empty?

What You'll Learn