Why Weight the Coordinates?
A point in the Hilbert cube has infinitely many coordinates, each lying in the unit interval. To measure the distance between two such points, a sum of coordinate differences is a natural idea, but the unweighted sum may be infinite. Giving later coordinates progressively smaller weights fixes this problem: the first few coordinates have substantial influence, while the total contribution from all remaining coordinates can be made small. We will prove that the resulting distance defines a metric and that its open sets are exactly those of the product topology.
Throughout, \(\mathbb{N}=\{1,2,\dots\}\), and \([0,1]\) has its usual subspace topology from \(\mathbb{R}\). The set \([0,1]^{\mathbb{N}}\) consists of all sequences \(x=(x_n)_{n\in\mathbb{N}}\) with \(x_n\in[0,1]\). Its product topology has a basis made from restrictions on finitely many coordinates: a typical basic set is \(\{y:y_n\in V_n\text{ for }n\in F\}\), where \(F\subseteq\mathbb{N}\) is finite and each \(V_n\) is open in \([0,1]\). Coordinates outside \(F\) are unrestricted.
Each coordinate difference is at most \(1\), so the series is finite: \(0\le d(x,y)\le\sum_{n=1}^{\infty}2^{-n}=1\). The weights are chosen so that the tail after coordinate \(N\) is especially simple: $$ \sum_{n=N+1}^{\infty}2^{-n}=2^{-N}. $$ In particular, no matter how the coordinates after \(N\) vary, their total contribution to the distance is at most \(2^{-N}\).
The Weighted Formula Is a Metric
Proof. For all \(x,y\in Q\), every summand \(2^{-n}|x_n-y_n|\) is nonnegative, so \(d(x,y)\ge0\). Since \(|x_n-y_n|=|y_n-x_n|\) in every coordinate, \(d(x,y)=d(y,x)\). If \(x=y\), every summand is zero and \(d(x,y)=0\). Conversely, if \(d(x,y)=0\), each nonnegative summand must be zero: if one were positive, the full sum would be positive. Thus \(|x_n-y_n|=0\) for every \(n\), so \(x_n=y_n\) for every \(n\), and hence \(x=y\).
For the triangle inequality, let \(x,y,z\in Q\). In each coordinate, the triangle inequality in \(\mathbb{R}\) gives \(|x_n-z_n|\le |x_n-y_n|+|y_n-z_n|\). Multiplying by \(2^{-n}>0\), summing from \(n=1\) to any finite \(m\), and then letting \(m\) tend to infinity gives $$ d(x,z)\le d(x,y)+d(y,z). $$ The sums converge because each is bounded above by \(\sum_{n=1}^{\infty}2^{-n}=1\). Therefore \(d\) satisfies nonnegativity, definiteness, symmetry, and the triangle inequality, which are the metric axioms. \(\square\)
Worked Example: Computing a Distance
Let \(x_n=0\) for every \(n\). Let \(y_1=\frac12\), \(y_3=1\), and \(y_n=0\) for all other indices. Compute \(d(x,y)\).
Only coordinates \(1\) and \(3\) have nonzero differences. Their contributions are \(2^{-1}|0-\frac12|=\frac12\cdot\frac12=\frac14\) and \(2^{-3}|0-1|=\frac18\cdot1=\frac18\). Every other summand is zero. Therefore $$ d(x,y)=\frac14+\frac18=\frac38. $$ Check: reversing the points leaves both absolute differences unchanged, so the formula gives \(d(y,x)=\frac14+\frac18=\frac38\), as symmetry requires.
Worked Example: The Tail Bound Is Exact
Let \(N=2\). Define \(x_1=x_2=y_1=y_2=0\), and for every \(n\ge3\), set \(x_n=0\) and \(y_n=1\). The points agree in the first two coordinates, so only the tail contributes: $$ d(x,y)=\sum_{n=3}^{\infty}2^{-n}|0-1| =\sum_{n=3}^{\infty}2^{-n} =\frac{1/8}{1-1/2} =\frac14. $$ The geometric-series computation gives \(2^{-2}=\frac14\), exactly the general tail bound for \(N=2\). Thus the bound cannot be decreased for all pairs that agree in their first two coordinates.
More generally, if \(x_n=y_n\) for \(1\le n\le N\), then $$ d(x,y)=\sum_{n=N+1}^{\infty}2^{-n}|x_n-y_n| \le\sum_{n=N+1}^{\infty}2^{-n} =2^{-N}, $$ because \(|x_n-y_n|\le1\) in every coordinate. This is the quantitative reason distant coordinates can be left unrestricted when constructing a product neighborhood.
The Metric Gives the Product Topology
Proof. We compare neighborhoods at an arbitrary point \(x\in Q\). First take a basic product neighborhood \(P\) of \(x\). It restricts a finite set \(F\) of coordinates, and for each \(n\in F\) we have \(x_n\in V_n\), where \(V_n\) is open in \([0,1]\). If \(F\) is empty, \(P=Q\), which contains every metric ball centered at \(x\). Otherwise, openness in the subspace \([0,1]\) gives a number \(r_n>0\) such that \((x_n-r_n,x_n+r_n)\cap[0,1]\subseteq V_n\). Define \(\delta=\min_{n\in F}2^{-n}r_n\). This is positive because it is the minimum of finitely many positive numbers.
If \(d(x,y)<\delta\), then for each \(n\in F\), \(2^{-n}|x_n-y_n|\le d(x,y)<\delta\le2^{-n}r_n\). Dividing by \(2^{-n}>0\) gives \(|x_n-y_n|<r_n\), so \(y_n\in V_n\). Thus \(B_d(x,\delta)\subseteq P\): every basic product neighborhood of \(x\) contains a metric ball about \(x\).
For the reverse inclusion, let \(\varepsilon>0\). Choose \(N\in\mathbb{N}\) large enough that \(2^{-N}<\varepsilon/2\), and consider the basic product neighborhood $$ P=\{y\in Q:|y_n-x_n|<\varepsilon/2\text{ for }1\le n\le N\}. $$ Each restriction is open in \([0,1]\). For \(y\in P\), the contribution from the first \(N\) coordinates satisfies $$ \sum_{n=1}^{N}2^{-n}|x_n-y_n| <\frac{\varepsilon}{2}\sum_{n=1}^{N}2^{-n} <\frac{\varepsilon}{2}, $$ since \(\sum_{n=1}^{N}2^{-n}<1\). The remaining contribution is at most \(\sum_{n=N+1}^{\infty}2^{-n}=2^{-N}<\varepsilon/2\). Adding the two bounds gives \(d(x,y)<\varepsilon\), so \(P\subseteq B_d(x,\varepsilon)\).
We have shown that every product-open neighborhood contains a metric ball around each of its points, and every metric ball contains a product-open neighborhood of its center. Consequently, a set is open in one topology exactly when it is open in the other: at each of its points, it contains a neighborhood from either local basis. Thus the metric topology and product topology agree. \(\square\)
Worked Example: Comparing a Ball and a Product Neighborhood
Let \(x_n=\frac12\) for every \(n\). First, consider the product neighborhood \(P=\{y:|y_n-\frac12|<\frac18\text{ for }n=1,2,3\}\). For \(y\in P\), the first three coordinates contribute less than $$ \frac18\left(\frac12+\frac14+\frac18\right)=\frac{7}{64}. $$ The tail from \(n=4\) onward contributes at most \(\sum_{n=4}^{\infty}2^{-n}=2^{-3}=\frac18=\frac{8}{64}\). Therefore \(d(x,y)<\frac{7}{64}+\frac{8}{64}=\frac{15}{64}<\frac14\), so \(P\subseteq B_d(x,\frac14)\).
In the other direction, let \(R=\{y:|y_1-\frac12|<\frac14,\ |y_2-\frac12|<\frac14\}\). For \(y\in R\), the first two coordinates contribute less than \(\frac14(\frac12+\frac14)=\frac{3}{16}\), while the tail from \(n=3\) onward contributes at most \(\sum_{n=3}^{\infty}2^{-n}=\frac14\). Hence \(d(x,y)<\frac{3}{16}+\frac14=\frac{7}{16}<\frac12\), and \(R\subseteq B_d(x,\frac12)\). The two inclusions illustrate how finite coordinate restrictions control an initial part of the sum, while the weight tail controls everything else.
Check: the bounds use strict inequalities for the restricted coordinates, and the tail bounds use \(|x_n-y_n|\le1\). Thus the first total is strictly below \(\frac{15}{64}\), and the second is strictly below \(\frac{7}{16}\), as asserted.
A Common Pitfall and the Role of the Weights
Common pitfall. One might try to define the distance as the unweighted sum \(\sum_{n=1}^{\infty}|x_n-y_n|\). For the points \(x_n=0\) and \(y_n=1\) for every \(n\), this would give \(1+1+1+\cdots\), which diverges. It is therefore not a finite-valued distance on \(Q\). With the stated weights, the same pair has $$ d(x,y)=\sum_{n=1}^{\infty}2^{-n}|0-1| =\sum_{n=1}^{\infty}2^{-n} =1. $$ The weighted sum is finite, and the proof above verifies all the metric axioms, not just finiteness.
The weights also explain how the metric represents product neighborhoods. A product neighborhood only needs to restrict finitely many coordinates; it need not constrain every coordinate by the same amount. Conversely, a sufficiently small metric ball forces closeness in any chosen finite collection of coordinates. The tail estimate ensures that unrestricted coordinates cannot outweigh these finite restrictions. The metric is therefore not merely a way to assign distances: it provides a single numerical description of the product topology on this countable product.
Check Your Understanding
Use the definition of \(d\) and the finite-coordinate basis for the product topology in your answers.
- Suppose \(x_n=y_n\) for \(1\le n\le5\). What upper bound does the tail estimate give for \(d(x,y)\)?
- For \(x_n=0\) for all \(n\), and \(y_2=1\) with all other \(y_n=0\), compute \(d(x,y)\).
- Explain why the set \(\{y:|y_1-x_1|<r\}\), for \(r>0\), is open in the product topology.
- In the proof that metric balls contain product neighborhoods, why must \(N\) be chosen so that \(2^{-N}<\varepsilon/2\)?
- Give two points of \(Q\) for which the unweighted sum of coordinate differences diverges, and compute their weighted distance.