A Metric for the Hilbert Cube Product Topology

A Metric Built from Coordinate Distances

The Hilbert cube has infinitely many coordinates, so a neighborhood in its product topology restricts only finitely many coordinates at a time. A useful metric must reflect that feature: differences in early coordinates should matter more, while changes far out in the coordinate list should have progressively smaller influence. We will define such a metric and prove that its open balls generate exactly the product topology.

Write \(I=[0,1]\) with its usual subspace topology from \(\mathbb{R}\), and let \(Q=I^{\mathbb{N}}\) with the product topology. Thus a point \(x\in Q\) is a sequence \(x=(x_1,x_2,\ldots)\) with \(x_n\in I\). A basic product neighborhood of \(x\) restricts coordinates in some finite set \(F\subseteq\mathbb{N}\): for each \(n\in F\), it requires \(y_n\) to lie in a relatively open subset of \(I\) containing \(x_n\), and it places no restriction on the other coordinates.

The key feature is that coordinates far along the sequence have a small total possible contribution. For every \(N\in\mathbb{N}\), the geometric-series formula gives $$ \sum_{n=N+1}^{\infty}2^{-n}=2^{-N}. $$ Even if two points differ by the maximum amount, \(1\), in every coordinate after \(N\), their total distance contribution from those coordinates is at most \(2^{-N}\).

Why the Formula Is a Metric

Proof. We check the metric axioms. Each summand \(2^{-n}|x_n-y_n|\) is nonnegative, so \(d(x,y)\geq 0\). Symmetry follows from \(|x_n-y_n|=|y_n-x_n|\), giving \(d(x,y)=d(y,x)\). If \(x=y\), every summand is zero and \(d(x,y)=0\). Conversely, if \(d(x,y)=0\), then each nonnegative summand is at most the sum and hence is zero. Thus \(|x_n-y_n|=0\), so \(x_n=y_n\), for every \(n\); consequently \(x=y\).

For the triangle inequality, let \(x,y,z\in Q\). For every \(n\), the triangle inequality on \(\mathbb{R}\) gives \(|x_n-z_n|\leq |x_n-y_n|+|y_n-z_n|\). Multiplying by \(2^{-n}\) and summing the nonnegative terms yields $$ d(x,z)\leq d(x,y)+d(y,z). $$ All three series converge because each is bounded above by \(\sum_{n=1}^{\infty}2^{-n}=1\). Therefore the inequality is valid for these sums, and all metric axioms hold. \(\square\)

Comparing the Two Kinds of Neighborhood

For \(x\in Q\) and \(r>0\), write \(B_d(x,r)=\{y\in Q:d(x,y)<r\}\) for the open metric ball. We now prove both containments needed to identify the metric topology with the product topology. One direction controls the finitely many coordinates constrained by a product neighborhood. The other makes the total contribution from the unrestricted tail small.

Proof. First let \(x\in Q\), and let \(U\) be a basic product neighborhood of \(x\). It restricts a finite set \(F\) of coordinates, with the requirement \(y_n\in V_n\) for each \(n\in F\), where \(V_n\) is relatively open in \(I\) and contains \(x_n\). For each \(n\in F\), relative openness gives a number \(\delta_n>0\) such that $$ \{t\in I:|t-x_n|<\delta_n\}\subseteq V_n. $$ If \(F\) is empty, then \(U=Q\), which contains every ball about \(x\). Otherwise, since \(F\) is finite and nonempty, the number \(r=\min_{n\in F}(2^{-n}\delta_n)\) is positive. If \(d(x,y)<r\), then for each \(n\in F\), $$ 2^{-n}|x_n-y_n|\leq d(x,y)<r\leq 2^{-n}\delta_n. $$ Dividing by \(2^{-n}>0\) gives \(|x_n-y_n|<\delta_n\), so \(y_n\in V_n\). Thus \(B_d(x,r)\subseteq U\).

Conversely, let \(\varepsilon>0\). Choose \(N\in\mathbb{N}\) so that \(2^{-N}<\varepsilon/2\), and put \(\delta=\varepsilon/2\). Consider the basic product neighborhood $$ V=\{y\in Q: |y_n-x_n|<\delta\text{ for }1\leq n\leq N\}. $$ For \(y\in V\), the contribution from the first \(N\) coordinates satisfies $$ \sum_{n=1}^{N}2^{-n}|x_n-y_n| < \delta\sum_{n=1}^{N}2^{-n} < \delta. $$ The remaining contribution is bounded by the full tail: $$ \sum_{n=N+1}^{\infty}2^{-n}|x_n-y_n| \leq\sum_{n=N+1}^{\infty}2^{-n}=2^{-N}<\frac{\varepsilon}{2}. $$ Adding the two estimates gives \(d(x,y)<\delta+\varepsilon/2=\varepsilon\). Therefore \(V\subseteq B_d(x,\varepsilon)\).

We have shown that every basic product neighborhood of \(x\) contains a metric ball about \(x\), and every metric ball about \(x\) contains a basic product neighborhood of \(x\). Since open sets in either topology are unions of neighborhoods of their points, the two topologies agree. \(\square\)

Why the Weights Matter

The metric makes the infinite product manageable without ignoring any coordinate. Every coordinate has a positive weight, so a difference in even a very late coordinate contributes to the distance. But its contribution can be small, which matches the product topology’s practice of controlling only finitely many coordinates in a basic neighborhood.

Common pitfall. Replacing the weights by \(1\) and writing \(\sum_{n=1}^{\infty}|x_n-y_n|\) does not define a finite-valued metric on all of \(Q\). For \(x=(0,0,\ldots)\) and \(y=(1,1,\ldots)\), its partial sum through coordinate \(N\) is \(\sum_{n=1}^{N}1=N\), which grows without bound as \(N\) increases. The weighted formula instead gives \(d(x,y)=\sum_{n=1}^{\infty}2^{-n}=1\).

Another tempting but incorrect inference is that \(d(x,y)<\varepsilon\) forces \(|x_n-y_n|<\varepsilon\) for every \(n\). The correct consequence from the \(n\)-th term is \(|x_n-y_n|<2^n\varepsilon\), which becomes less restrictive as \(n\) increases. For example, if \(y\) differs from the zero sequence only at coordinate \(10\), where \(y_{10}=1\), then \(d(x,y)=2^{-10}=1/1024\) even though \(|x_{10}-y_{10}|=1\). The metric’s small weights are exactly why it can represent the product topology rather than a topology that uniformly controls all coordinates.

This metric is useful because it lets us use metric language and techniques on the Hilbert cube while preserving its product topology. The proof depends on two complementary facts: finitely many coordinate restrictions can be enforced by a sufficiently small ball, and the total effect of all coordinates beyond a sufficiently large index can be made arbitrarily small.