Connected Boxes in Euclidean Space

Boxes and Connectedness

A box in Euclidean space may include all of its boundary, none of it, or only some of it. These choices affect whether the box is open or closed, but they do not prevent a nonempty box from being connected. The reason is that each coordinate interval is connected, and connectedness is preserved by products.

We use the standard topology on \(\mathbb{R}\) and the Euclidean topology on \(\mathbb{R}^n\), where \(n\in\mathbb{N}\). Each interval below has the subspace topology inherited from \(\mathbb{R}\).

Definition: A box in \(\mathbb{R}^n\) is a product \(B=I_1\times\cdots\times I_n\), where each \(I_i\subseteq\mathbb{R}\) is an interval. The intervals may be open, closed, half-open, unbounded, or singletons. A topological space is connected if it cannot be written as the union of two disjoint, nonempty open subsets of that space. Here, when considering a box as a space, its topology is the subspace topology from \(\mathbb{R}^n\).

For example, \((0,1)\times[2,4]\) is a box in \(\mathbb{R}^2\). Its first coordinate is restricted to values strictly between \(0\) and \(1\); its second coordinate includes both \(2\) and \(4\). Connectedness is a property of the topology on the box, not a claim that every boundary point is included.

Products and the Euclidean Topology

To apply a product result, we first relate the product topology on the coordinate intervals to the topology inherited from \(\mathbb{R}^n\). The product topology has a basis consisting of products of open sets in the individual coordinates.

Lemma: For \(I_1,\ldots,I_n\subseteq\mathbb{R}\), the product topology on \(I_1\times\cdots\times I_n\) agrees with the subspace topology inherited from Euclidean \(\mathbb{R}^n\).

Proof. First, the product topology on \(\mathbb{R}^n\) agrees with its Euclidean topology. A product of open intervals is Euclidean open: at any point in the product, choose a positive radius smaller than the distance from each coordinate to the endpoints of its interval. The Euclidean ball of that radius is contained in the product. Conversely, let \(x\in\mathbb{R}^n\) and let \(B(x,r)\) be a Euclidean ball with \(r>0\). The product of the intervals \((x_i-r/(2\sqrt{n}),x_i+r/(2\sqrt{n}))\), for \(i=1,\ldots,n\), contains \(x\) and lies inside \(B(x,r)\): for a point \(y\) in that product, \(\|y-x\|<\sqrt{n}\,r/(2\sqrt{n})=r/2<r\). Thus the two topologies on \(\mathbb{R}^n\) have the same basis neighborhoods. Intersecting these product neighborhoods with \(I_1\times\cdots\times I_n\) gives products of relatively open sets in the \(I_i\). Those products form a basis for the product topology on the intervals, so the product and subspace topologies agree. \(\square\)

We now use the standard product theorem from the earlier material on arbitrary products of connected spaces: a product of connected spaces, with the product topology, is connected. Its hypotheses require each factor to be connected; for a nonempty product, each factor must also be nonempty.

Theorem: Every nonempty box \(B=I_1\times\cdots\times I_n\) in Euclidean \(\mathbb{R}^n\) is connected, for every choice of open, closed, half-open, or unbounded interval factors.

Proof. Suppose \(B\ne\varnothing\). Then every factor \(I_i\) is nonempty. Each \(I_i\), as an interval in \(\mathbb{R}\), is connected by the theorem that every interval in \(\mathbb{R}\) is connected. The product theorem therefore says that \(I_1\times\cdots\times I_n\) is connected in the product topology. By the lemma, that topology is exactly the subspace topology inherited from Euclidean \(\mathbb{R}^n\). Hence \(B\) is connected as a subspace of \(\mathbb{R}^n\). If a factor is empty, then \(B=\varnothing\); the empty space is connected because it cannot be split into two nonempty sets. \(\square\)

Worked Examples: Open, Closed, and Mixed Boxes

Worked Example: An Open Box

Consider \(B=(0,2)\times(1,3)\subseteq\mathbb{R}^2\). Each coordinate interval is nonempty and connected, so the theorem gives that \(B\) is connected. It is also open in \(\mathbb{R}^2\): for any \((x,y)\in B\), the four numbers \(x\), \(2-x\), \(y-1\), and \(3-y\) are positive. Let \(r\) be half the minimum of those four numbers. Then \(r>0\), and every point within Euclidean distance \(r\) of \((x,y)\) remains strictly between the corresponding endpoints in both coordinates. Thus \(B((x,y),r)\subseteq B\).

Check: At \((1,2)\), all four distances to the coordinate endpoints are \(1\), so \(r=1/2\) works. Every point in \(B((1,2),1/2)\) has each coordinate within \(1/2\) of \(1\) or \(2\), respectively, and so belongs to \((0,2)\times(1,3)\).

Worked Example: A Closed Box

Consider \(C=[-1,1]\times[2,5]\subseteq\mathbb{R}^2\). Its factors are nonempty intervals, so \(C\) is connected. It is closed as well: if a point \((x,y)\) lies outside \(C\), then either \(x\notin[-1,1]\) or \(y\notin[2,5]\). In the first case, the closed interval \([-1,1]\) has an open neighborhood of \(x\) disjoint from it; that neighborhood, taken in the first coordinate, gives a neighborhood of \((x,y)\) disjoint from \(C\). In the second case, the same argument uses the second coordinate. Therefore the complement of \(C\) is open.

Check: The point \((0,3)\) belongs to \(C\), since \(-1\leq0\leq1\) and \(2\leq3\leq5\). The point \((2,3)\) is outside \(C\), since \(2>1\). An open interval around \(2\) that misses \([-1,1]\), used in the first coordinate, gives a Euclidean neighborhood of \((2,3)\) disjoint from \(C\).

Worked Example: A Mixed-Endpoint Box

Let \(D=[0,1)\times(2,4]\subseteq\mathbb{R}^2\). Both factors are nonempty intervals, so \(D\) is connected. It is neither open nor closed. It is not open because \((0,3)\in D\), but every Euclidean ball centered there contains points with first coordinate less than \(0\), which are outside \(D\). More explicitly, for every \(r>0\), the point \((- \min(r/2,1/2),3)\) is within distance \(r\) of \((0,3)\) and is not in \(D\).

It is not closed because \((1,3)\notin D\), while the points \((1-1/m,3)\in D\), for integers \(m\geq2\), converge to \((1,3)\). A closed subset of a metric space contains the limit of every convergent sequence of its points: if the limit were outside the closed set, its open complement would contain a neighborhood of the limit and eventually contain the sequence terms, a contradiction.

Check: For \(m=2\), the point is \((1/2,3)\in D\); as \(m\) increases, \(1-1/m\) tends to \(1\), while the second coordinate remains \(3\). Thus the excluded endpoint is indeed approached by points of \(D\).

How Endpoint Choices Determine Openness and Closedness

Connectedness holds across all these examples. Openness and closedness, by contrast, depend on the coordinate intervals. For a nonempty box, each coordinate interval can be tested separately.

Proposition: A nonempty box \(B=I_1\times\cdots\times I_n\) is open in \(\mathbb{R}^n\) if and only if every \(I_i\) is open in \(\mathbb{R}\). It is closed in \(\mathbb{R}^n\) if and only if every \(I_i\) is closed in \(\mathbb{R}\).

Proof. Suppose first that every \(I_i\) is open. Each point of the box has a product neighborhood contained in the box; by the equality of product and Euclidean topologies, the box is open. Conversely, suppose \(B\) is open and choose any \(x\in B\). There is \(r>0\) such that \(B(x,r)\subseteq B\). Fix a coordinate \(i\). For every \(t\in\mathbb{R}\) with \(|t-x_i|<r\), the point obtained from \(x\) by replacing only its \(i\)-th coordinate by \(t\) lies in \(B(x,r)\), and hence \(t\in I_i\). Thus \(I_i\) contains an open interval around each of its points and is open.

Now suppose every \(I_i\) is closed, and let \(x\notin B\). Some coordinate \(x_i\notin I_i\). Since \(I_i\) is closed, \(\mathbb{R}\setminus I_i\) is open, so there is an open interval around \(x_i\) missing \(I_i\). Restricting that coordinate to this interval gives a neighborhood of \(x\) disjoint from \(B\). Hence the complement of \(B\) is open, and \(B\) is closed. Conversely, suppose \(B\) is closed. Fix \(i\), and fix points \(c_j\in I_j\) for all \(j\ne i\), which is possible because the box is nonempty. The map that sends \(t\in\mathbb{R}\) to the point with \(i\)-th coordinate \(t\) and other coordinates \(c_j\) is continuous. Its preimage of \(B\) is \(I_i\). A continuous preimage of a closed set is closed, so \(I_i\) is closed. This holds for every \(i\). \(\square\)

For a nondegenerate interval with finite endpoints \(a<b\), the endpoint test is direct: \((a,b)\) excludes both endpoints and is open; \([a,b]\) includes both and is closed; \([a,b)\) includes just the left endpoint and is neither; \((a,b]\) includes just the right endpoint and is neither. For rays, the same test applies at the finite endpoint, while an infinite end contributes no endpoint to include or exclude. The proposition applies these tests coordinate by coordinate. A singleton interval is closed but not open in \(\mathbb{R}\).

Thus, for a nonempty box with nondegenerate factors, it is open exactly when every finite endpoint is excluded, and closed exactly when every finite endpoint is included. A box may be both open and closed in special cases, such as \(\mathbb{R}^n\); the criteria do not say that open and closed are opposites. The empty box is also both open and closed, although the proposition above was stated for nonempty boxes.

A Common Pitfall: Mistaking a Boundary for a Separation

A missing edge or face does not by itself make a box disconnected. For instance, consider \(D=[0,1)\times[0,1]\), which is connected by the box theorem. A tempting attempted separation divides it into \(A=[0,1/2)\times[0,1]\) and \(E=[1/2,1)\times[0,1]\). These sets are disjoint, nonempty, and have union \(D\), but that is not enough for a separation: both parts must also be open in the subspace \(D\).

In fact, \(E\) is not open in \(D\). The point \((1/2,1/2)\) lies in \(E\). Every Euclidean neighborhood of it contains points \((1/2-\delta,1/2)\) for sufficiently small \(\delta>0\); those points lie in \(D\) but not in \(E\). So no relatively open neighborhood of \((1/2,1/2)\) is contained in \(E\). The proposed split is not a separation and cannot establish disconnectedness.

Takeaway: For a nonempty box, connectedness comes from connectedness of its interval factors and the product theorem. Endpoint inclusion determines whether the box is open or closed; it does not determine whether the box is connected.

Check Your Understanding

For each question, distinguish the topology of a box from the endpoint choices defining it.

Is \((0,1]\times[2,3)\) connected, open, closed, both open and closed, or neither open nor closed in \(\mathbb{R}^2\)?
Which hypothesis in the box connectedness theorem ensures that each coordinate factor is connected?
Explain why \([0,1]\times\{2\}\) is connected, and determine whether it is open or closed in \(\mathbb{R}^2\).
For a nonempty box, prove that if one coordinate factor is not closed, then the box is not closed.
Why does dividing a box into two disjoint nonempty pieces fail to prove disconnectedness unless both pieces are relatively open?