The Hilbert Cube and Its Points
The Hilbert cube is a familiar space built from infinitely many simple pieces: each coordinate is a real number between zero and one. Its topology records closeness by restricting finitely many coordinates at a time. Understanding this local structure is essential before asking what other topological properties the cube has.
Throughout, \(I=[0,1]\) has the subspace topology inherited from \(\mathbb{R}\) with its standard topology. Let \(\mathbb{N}=\{1,2,\dots\}\). The product \(I^{\mathbb{N}}\) is given the product topology, not the box topology.
The sequence notation is simply a way to record the value of the function \(x\) at each natural number: \(x_n=x(n)\). Thus \(Q\) contains all sequences of real numbers whose entries lie in \([0,1]\). It is not restricted to sequences that are eventually constant, eventually zero, or convergent. For example, the coordinates can alternate between zero and one indefinitely.
For each \(n\in\mathbb{N}\), the coordinate projection \(\pi_n:Q\to I\) is defined by \(\pi_n(x)=x_n\). A condition on one coordinate, such as \(x_4\in(1/3,2/3)\), is therefore the preimage under \(\pi_4\) of a subset of \(I\). The product topology is designed so that these coordinate conditions, and their finite intersections, form the basic open sets.
Basic Neighborhoods in the Product Topology
“Open in \(I\)” means open for the subspace topology. For example, \([0,1/4)\) is open in \(I\), since \([0,1/4)=I\cap(-1,1/4)\). Likewise, \((3/4,1]\) is open in \(I\), since it is \(I\cap(3/4,2)\). These relative neighborhoods matter at the endpoints: a neighborhood of \(0\) in \(I\) need not extend to negative real numbers.
Proof. We verify the basis conditions. First, every \(x\in Q\) belongs to \(Q=\prod_{n\in\mathbb{N}}I\), which is basic because each coordinate set is open in \(I\) and no coordinates are restricted. Second, suppose \(x\) belongs to two basic open sets \(B=\prod_n U_n\) and \(C=\prod_n V_n\). Their intersection is \(\prod_n(U_n\cap V_n)\). Each \(U_n\cap V_n\) is open in \(I\), and it equals \(I\) except possibly at coordinates restricted by \(B\) or \(C\). There are only finitely many such coordinates, so \(B\cap C\) is basic and contains \(x\). These two conditions show that the stated sets form a basis. The product topology is the topology generated by this basis, so each open set containing \(x\) contains a basis element that contains \(x\). \(\square\)
This theorem gives a practical neighborhood test. To describe a particular basic neighborhood of \(x\), choose a finite set \(F\subseteq\mathbb{N}\), and for each \(n\in F\) choose an open set \(U_n\subseteq I\) containing \(x_n\). The resulting neighborhood is \(\{y\in Q:y_n\in U_n\text{ for every }n\in F\}\). There are no conditions on \(y_n\) when \(n\notin F\).
Worked Example: A Neighborhood of a Specified Point
Let \(x\in Q\) have coordinates \(x_1=0\), \(x_2=1/2\), \(x_3=1\), and \(x_n=1/3\) for \(n\ge4\). Construct a basic open neighborhood of \(x\) that restricts its first three coordinates.
In the first coordinate, choose \(U_1=[0,1/4)\). It is open in \(I\), and \(x_1=0\in U_1\). In the second, choose \(U_2=(1/3,2/3)\). This is open in \(I\), and \(1/3<1/2<2/3\), so \(x_2\in U_2\). In the third, choose \(U_3=(3/4,1]\). It is open in \(I\), and \(3/4<1\le1\), so \(x_3\in U_3\). Let \(U_n=I\) for every \(n\ge4\). The neighborhood is \(B=U_1\times U_2\times U_3\times\prod_{n\ge4}I\).
Thus \(y\in B\) exactly when \(0\le y_1<1/4\), \(1/3<y_2<2/3\), and \(3/4<y_3\le1\); all coordinates \(y_n\) for \(n\ge4\) may be any values in \([0,1]\). The point \(x\) satisfies all three restrictions, so \(x\in B\). As a check, the point \(y\) with \(y_1=1/8\), \(y_2=1/2\), \(y_3=1\), and \(y_n=0\) for \(n\ge4\) lies in \(B\): \(0\le1/8<1/4\), \(1/3<1/2<2/3\), and \(3/4<1\le1\). The unrestricted later coordinates do not affect membership.
Coordinate Projections and Cylinders
Proof. A function is continuous if the preimage of every open set in its codomain is open in its domain. Let \(V\) be any open subset of \(I\). By the definition of the coordinate projection, \(\pi_k^{-1}(V)=\{x\in Q:x_k\in V\}\). This is the basic open set whose \(k\)-th coordinate set is \(V\) and whose coordinate sets at all \(n\ne k\) are \(I\). It is open in \(Q\). This also applies when \(V=\varnothing\), whose preimage is empty, and when \(V=I\), whose preimage is all of \(Q\). Therefore \(\pi_k\) is continuous. \(\square\)
A set obtained by restricting finitely many coordinates is often called a cylinder set: coordinates outside the chosen finite set do not enter its definition. This description is useful whether the restricted coordinate sets are open or not; the cylinder is a basic open set specifically when each restricted set is open in \(I\).
Worked Example: An Open Set Specified by Two Coordinates
Consider the set \(O=\{x\in Q:x_2\in[0,1/3),\ x_5\in(2/3,1]\}\). Decide whether it is open in \(Q\).
The interval \([0,1/3)\) is open in \(I\), because it equals \(I\cap(-1,1/3)\). The interval \((2/3,1]\) is open in \(I\), because it equals \(I\cap(2/3,2)\). Define \(U_2=[0,1/3)\), \(U_5=(2/3,1]\), and \(U_n=I\) for every \(n\ne2,5\). Then \(O=\prod_{n\in\mathbb{N}}U_n\). Only coordinates 2 and 5 are restricted, so this is a basic open set and hence is open.
For a membership check, take \(x_2=1/4\), \(x_5=3/4\), and set every other coordinate to zero. Then \(0\le1/4<1/3\) and \(2/3<3/4\le1\), so \(x\in O\). If instead \(x_2=1/3\), then \(x\notin O\), because the upper endpoint \(1/3\) is excluded by \([0,1/3)\). This check highlights the strictness of the first upper bound.
The proposition also explains the role of single-coordinate tests: every open set in \(I\) pulls back to an open coordinate cylinder in \(Q\). Finite intersections of such cylinders give the basic neighborhoods. The product topology allows a neighborhood to test several chosen coordinates at once, but requires only finitely many tests in any one basic neighborhood.
Why Finitely Many Restrictions Matter
The requirement that all but finitely many coordinate sets equal \(I\) is not a technical detail. It distinguishes the product topology from a topology whose basic sets could restrict infinitely many coordinates simultaneously. For example, a condition imposed in every coordinate may define a set, but it does not follow that the set is open in the product topology. The basis theorem provides a direct way to check: at each point of an open set, some finite-coordinate basic neighborhood must fit inside it.
Worked Example: An Infinite Coordinate Condition Is Not Open
Let \(A=\{x\in Q:x_n<1/2\text{ for every }n\in\mathbb{N}\}\). We show that \(A\) is not open in \(Q\).
The point \(z=(0,0,\ldots)\) belongs to \(A\), since \(z_n=0<1/2\) for every \(n\). Suppose \(A\) were open. The basis theorem would then give a basic open set \(B\) with \(z\in B\subseteq A\). Because \(B\) restricts only finitely many coordinates, there is an index \(k\) whose coordinate is unrestricted: \(U_k=I\) in the product description of \(B\).
Define \(y\in Q\) by \(y_k=1/2\) and \(y_n=0\) for every \(n\ne k\). The unrestricted \(k\)-th coordinate and the fact that \(0\in U_n\) at each restricted coordinate imply \(y\in B\). But \(y_k=1/2\) does not satisfy the strict inequality \(y_k<1/2\), so \(y\notin A\). This contradicts \(B\subseteq A\). Therefore \(A\) is not open.
Check: The failure is exactly at a coordinate left unrestricted by any proposed basic neighborhood. It does not depend on which finite set of coordinates that neighborhood restricts: \(\mathbb{N}\) has an index outside every finite subset.
Common pitfall. A tempting but incorrect argument says that \(A\) is open because each one-coordinate condition \(x_n<1/2\) defines an open set in \(Q\), and \(A\) is the intersection of those sets. Indeed, each set \(A_n=\{x\in Q:x_n<1/2\}\) is open by the projection result. However, the topology guarantees that finite intersections of open sets are open, not arbitrary infinite intersections. Here \(A=\bigcap_{n\in\mathbb{N}}A_n\), and the explicit neighborhood argument above verifies that this particular infinite intersection is not open.
A Practical Description of Local Structure
To work with an open neighborhood in the Hilbert cube, first identify the finitely many coordinates that matter. At each of them, choose a neighborhood in \(I\) of the coordinate of the point. All other coordinates remain free to range over the whole interval. Conversely, any open neighborhood of a point contains at least one neighborhood of this form. This gives both a way to construct open sets and a way to disprove openness: if every finite-coordinate neighborhood around a point includes a point outside the proposed set, that set is not open.
| Feature | Product-topology description |
|---|---|
| A point of \(Q\) | A sequence \(x=(x_1,x_2,\ldots)\) with \(x_n\in[0,1]\) for every \(n\) |
| A basic open neighborhood | Open restrictions in finitely many coordinates; all other coordinates range over \(I\) |
| Endpoint restriction | Sets such as \([0,a)\) and \((b,1]\) can be open in the subspace \(I\) |
| Coordinate projection | \(\pi_k(x)=x_k\), a continuous map from \(Q\) to \(I\) |
| Infinite coordinate condition | It is not automatically open; test whether a finite-coordinate neighborhood fits inside it |
Check Your Understanding
Use the finite-coordinate basis to justify each answer.
- Describe a basic open neighborhood of \(x\in Q\) that restricts only coordinates 1 and 4, and state precisely what happens at all other coordinates.
- Explain why \(\{x\in Q:x_3\in[0,1/2)\}\) is open, including why the interval used is open in \(I\).
- Is \(\{x\in Q:x_n\le1/2\text{ for every }n\in\mathbb{N}\}\) open? Test the point \((0,0,\ldots)\) using an arbitrary basic neighborhood.
- For an open set \(V\subseteq I\), write \(\pi_7^{-1}(V)\) as a product whose coordinate sets are specified.
- Give an example of two points in the same basic neighborhood even though they differ in infinitely many coordinates.