Compactness of the Hilbert Cube

The Hilbert Cube and Its Product Topology

The Hilbert cube is an infinite product of compact intervals. Its compactness is an important illustration of how compactness can persist in a product with infinitely many coordinates: the relevant theorem is Tychonoff’s theorem, not a direct finite-dimensional argument. The topology matters. Throughout, the Hilbert cube carries the product topology, in which a basic open set restricts only finitely many coordinates.

Definition: The Hilbert cube is $$ H=\prod_{n\in\mathbb{N}}[0,1/n], $$ where each interval has its usual subspace topology from $\mathbb{R}$, and $H$ has the product topology. A point $x\in H$ is a sequence $x=(x_1,x_2,\ldots)$ with $0\leq x_n\leq 1/n$ for every $n$.

In the product topology, a basis consists of sets of the form $\prod_{n\in\mathbb{N}}U_n$, where each $U_n$ is open in $[0,1/n]$ and $U_n=[0,1/n]$ for all but finitely many $n$. Such a set is called a basic cylinder: membership is determined by restrictions on a finite set of coordinates. An arbitrary open set is a union of basic cylinders, and that union need not be finite.

For each $n$, the interval $[0,1/n]$ is compact in its usual topology by the Heine–Borel theorem: a closed and bounded interval in $\mathbb{R}$ is compact. The point $(0,0,\ldots)$ belongs to $H$, so the product is nonempty as well.

Compactness from Tychonoff’s Theorem

Theorem (Compactness of the Hilbert cube): The Hilbert cube $H=\prod_{n\in\mathbb{N}}[0,1/n]$, with the product topology, is compact.

Proof. Tychonoff’s theorem states that a product of compact topological spaces, equipped with the product topology, is compact. For every $n\in\mathbb{N}$, the factor $[0,1/n]$ is compact by Heine–Borel. Therefore every factor in the product defining $H$ is compact. Applying Tychonoff’s theorem to this family of factors gives that $H$ is compact. By the definition of compactness, every open cover of $H$ has a finite subcover. $\square$

The proof uses the product topology hypothesis in Tychonoff’s theorem. The theorem does not provide a recipe for selecting a finite subcover from a particular cover; explicit covers can still require their own analysis. The examples below show how finite-coordinate restrictions can make that analysis concrete.

Worked Example: A Countable Cover with a Finite Subcover

Let $x_1$ denote the first coordinate of $x\in H$. Consider $$ V=\{x\in H:x_1<1/3\},\qquad U_k=\{x\in H:x_1>1/(k+3)\}\quad(k\in\mathbb{N}). $$ Each set is open: it is the inverse image of an open subset of $[0,1]$ under the first-coordinate projection. Thus $\{V\}\cup\{U_k:k\in\mathbb{N}\}$ is an open cover candidate.

To check that it covers, take any $x\in H$. If $x_1<1/3$, then $x\in V$. If $x_1\geq1/3$, then $$ x_1\geq1/3>1/4=1/(1+3), $$ so $x\in U_1$. Hence $V\cup U_1=H$, and the two sets $V,U_1$ are a finite subcover of the countable family.

Check: at the boundary value $x_1=1/3$, the strict inequality defining $V$ fails, but $1/3>1/4$, so $U_1$ contains the point. At $x_1=0$, $V$ contains it. The endpoints therefore cause no gap in the cover.

Worked Example: Four Cylinders Cover the Cube

Rescale the first two coordinates by setting $t_1=x_1$ and $t_2=2x_2$. Since $x_1\in[0,1]$ and $x_2\in[0,1/2]$, both $t_1$ and $t_2$ lie in $[0,1]$. In $[0,1]$, let $L=[0,2/3)$ and $R=(1/3,1]$. These are relatively open: $L=(-1,2/3)\cap[0,1]$ and $R=(1/3,2)\cap[0,1]$. Also $L\cup R=[0,1]$.

For $A,B\in\{L,R\}$, define the cylinder $$ C_{A,B}=\{x\in H:t_1\in A,\ t_2\in B\}. $$ Each cylinder is open because it restricts only the first two coordinates to open sets. The four cylinders cover $H$: for each of $t_1,t_2$, choose $L$ if the value is less than $2/3$, and choose $R$ if it is at least $2/3$. In the second case the value is greater than $1/3$, so it belongs to $R$; in the first case it belongs to $L$. The resulting pair of choices gives one of the four cylinders containing $x$.

For a specific check, take a point with $x_1=1/2$ and $x_2=1/6$, and choose any permitted values for its remaining coordinates. Then $t_1=1/2>1/3$, so $t_1\in R$, while $t_2=2(1/6)=1/3<2/3$, so $t_2\in L$. Thus the point lies in $C_{R,L}$. The four cylinders themselves form a finite subcover, since they cover and there are exactly four of them.

Open Sets Can Involve Infinitely Many Coordinates

Worked Example: An Open Set That Is Not a Single Finite-Coordinate Cylinder

For each $n\in\mathbb{N}$, let $$ C_n=\{x\in H:x_n<1/(2n)\},\qquad O=\bigcup_{n\in\mathbb{N}}C_n. $$ Each $C_n$ is open: it restricts the $n$th coordinate to the relatively open set $[0,1/(2n))$ in $[0,1/n]$. Since a union of open sets is open, $O$ is open.

However, membership in $O$ is not determined by any one fixed finite set of coordinates. To verify this, let $F\subseteq\mathbb{N}$ be any finite set and choose $j\in\mathbb{N}\setminus F$. Define $z_j=0$ and $z_i=1/i$ for $i\ne j$. Then $z\in C_j$, because $0<1/(2j)$, so $z\in O$. Let $y_i=1/i$ for every $i$. For each $i$, the inequality $1/i<1/(2i)$ is false, so $y\notin O$. Yet $z_i=y_i$ for every $i\in F$, since $j\notin F$. Thus agreement on an arbitrary prescribed finite set of coordinates does not always decide membership in $O$.

This does not contradict the basis description of the product topology. Every point of an open set has a basic-cylinder neighborhood contained in that open set; an open set itself may be a union of cylinders involving different coordinates. The conclusion is about a single finite set of coordinates deciding membership everywhere, not about whether $O$ is open.

Compactness and Closed Families

Corollary: If $\{F_\alpha\}_{\alpha\in J}$ is a family of closed subsets of $H$ such that every finite subfamily has nonempty intersection, then $\bigcap_{\alpha\in J}F_\alpha\ne\varnothing$.

Proof. Suppose, to the contrary, that $\bigcap_{\alpha\in J}F_\alpha=\varnothing$. Taking complements in $H$, the open sets $H\setminus F_\alpha$ cover $H$: for each $x\in H$, failure of $x$ to belong to the intersection means that $x\notin F_\alpha$ for at least one index $\alpha$. Compactness of $H$ gives a finite subcover $H\setminus F_{\alpha_1},\ldots,H\setminus F_{\alpha_m}$. Their union is $H$, so $$ F_{\alpha_1}\cap\cdots\cap F_{\alpha_m} =H\setminus\bigl((H\setminus F_{\alpha_1})\cup\cdots\cup(H\setminus F_{\alpha_m})\bigr) =\varnothing. $$ This contradicts the assumption that every finite subfamily has nonempty intersection. Therefore the full intersection is nonempty. $\square$

Common pitfall. A tempting but incorrect calculation is: “Every open set in the product topology restricts finitely many coordinates; therefore every open set is determined by finitely many coordinates.” The first statement describes basic open sets, not all open sets. The set $O$ in the preceding example is open because it is a union of the open cylinders $C_n$, but for every finite $F$ we constructed $z,y$ agreeing on $F$ with $z\in O$ and $y\notin O$. The correction is to distinguish a basic cylinder from an arbitrary union of basic cylinders. Compactness applies to covers by arbitrary open sets, not just to covers whose members are single cylinders.

The role of compact factors is also essential to the stated application. For example, $\mathbb{R}^{\mathbb{N}}$ with the product topology is not compact. The sets $W_n=\{x\in\mathbb{R}^{\mathbb{N}}:|x_1|<n\}$, for $n\in\mathbb{N}$, are open and cover the product: for any point, its first coordinate is a real number, so its absolute value is less than some positive integer. But a finite selection has a largest index $N$, and its union is just $W_N$. The point with first coordinate $N$ and all other coordinates zero is not in $W_N$, because $|N|<N$ is false. Thus this cover has no finite subcover. This example does not challenge Tychonoff’s theorem; its factors $\mathbb{R}$ are not compact.

Check Your Understanding

Use the product topology and the strict inequalities in each definition when checking membership.

Why is each factor $[0,1/n]$ compact, and which theorem turns that fact into compactness of $H$?
For the cover in the first worked example, check directly which set contains a point with $x_1=1/3$.
Verify that $L=[0,2/3)$ and $R=(1/3,1]$ cover $[0,1]$, including the values $1/3$ and $2/3$.
Explain why the union $O=\bigcup_{n\in\mathbb{N}}\{x:x_n<1/(2n)\}$ is open even though no fixed finite set of coordinates determines membership in $O$.
State the finite intersection conclusion for a family of closed subsets of $H$ whose every finite subfamily has nonempty intersection.

Compactness of the Hilbert Cube

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