Pulling Back the Hilbert Cube Metric

From an Embedding to a Metric

A topological embedding does more than place one space inside another: it identifies the original topology with the topology on its image. If the containing space already has a metric, that metric can be transferred back along the embedding. The key questions are whether the transferred distance is genuinely a metric and whether its open sets are exactly the original open sets.

Let \(\mathbb{N}=\{1,2,\dots\}\), and give \([0,1]\) its usual topology. The Hilbert cube is \(Q=[0,1]^{\mathbb{N}}\), equipped with the product topology. A point \(u\in Q\) has coordinates \(u=(u_1,u_2,\dots)\). We use the weighted distance

The series is finite because \(0\le |u_n-v_n|\le 1\) and \(\sum_{n=1}^{\infty}2^{-n}=1\). We first verify both that this formula is a metric and that it induces the product topology.

The Metric and Topology of the Hilbert Cube

Proof. Every term in the series defining \(d_Q\) is nonnegative, so \(d_Q(u,v)\ge 0\). If \(u=v\), every coordinate difference is zero and \(d_Q(u,v)=0\). Conversely, if \(d_Q(u,v)=0\), each nonnegative term \(2^{-n}|u_n-v_n|\) must be zero. Since \(2^{-n}>0\), this gives \(u_n=v_n\) for every \(n\), hence \(u=v\). Symmetry follows from \(|u_n-v_n|=|v_n-u_n|\). For \(u,v,w\in Q\), the real triangle inequality gives \(|u_n-w_n|\le |u_n-v_n|+|v_n-w_n|\) for each \(n\). Multiplying by \(2^{-n}\), summing, and using convergence of the series gives \(d_Q(u,w)\le d_Q(u,v)+d_Q(v,w)\). Thus \(d_Q\) is a metric.

To compare the topologies, a basic product neighborhood of \(u\) restricts finitely many coordinates: it contains a set of the form \(\{v\in Q: |v_i-u_i|<\varepsilon_i\text{ for every }i\in F\}\), where \(F\subseteq\mathbb{N}\) is finite and each \(\varepsilon_i>0\). If \(F\ne\varnothing\), set \(r=\min_{i\in F}2^{-i}\varepsilon_i\). If \(d_Q(u,v)<r\), then \(2^{-i}|u_i-v_i|\le d_Q(u,v)<r\le 2^{-i}\varepsilon_i\) for each \(i\in F\), so \(|u_i-v_i|<\varepsilon_i\). Therefore a metric ball about \(u\) lies in the given product neighborhood. If \(F=\varnothing\), the basic neighborhood is all of \(Q\), so the inclusion is automatic.

For the other direction, fix \(r>0\). Choose \(N\in\mathbb{N}\) so that \(2^{-N}<r/2\), and let \(\delta=r/2\). The product neighborhood restricting the first \(N\) coordinates by \(|v_i-u_i|<\delta\) lies in the \(d_Q\)-ball of radius \(r\): its first \(N\) terms sum to less than \(\delta\sum_{i=1}^{N}2^{-i}<r/2\), while its remaining terms sum to at most \(\sum_{i=N+1}^{\infty}2^{-i}=2^{-N}<r/2\). Consequently \(d_Q(u,v)<r\). Each metric ball therefore contains a product neighborhood of its center, and each basic product neighborhood contains a metric ball of its center. The topologies agree. \(\square\)

The weights have two jobs: they make the infinite sum finite, and they make the contribution of all coordinates after \(N\) as small as \(2^{-N}\). The latter fact lets finitely many coordinate restrictions control a metric ball.

Pulling Back the Distance

The embedding theorem for regular second-countable spaces, established in earlier material, gives the needed map into \(Q\). In the form used here, it says that if \((X,\mathcal{T})\) is regular and second-countable, then there is a topological embedding \(e:X\to Q\). We do not need to reconstruct its coordinate functions here; we use the embedding it supplies and transfer \(d_Q\) along that map.

Proof. If \(X=\varnothing\), the empty function \(d_X:\varnothing\times\varnothing\to[0,\infty)\) satisfies the metric axioms vacuously, and the unique topology on \(\varnothing\) is its metric topology. Now suppose \(X\ne\varnothing\). Nonnegativity and symmetry of \(d_X\) follow from those of \(d_Q\). If \(d_X(x,y)=0\), then \(d_Q(e(x),e(y))=0\), so \(e(x)=e(y)\) because \(d_Q\) is a metric. Injectivity of \(e\) gives \(x=y\). Also \(d_X(x,x)=d_Q(e(x),e(x))=0\). For \(x,y,z\in X\), the triangle inequality for \(d_Q\) gives \(d_X(x,z)\le d_X(x,y)+d_X(y,z)\). Thus \(d_X\) is a metric.

For any \(x\in X\) and \(r>0\), its \(d_X\)-ball is exactly \(e^{-1}(B_{d_Q}(e(x),r))\). Since \(e\) is continuous, this preimage is open in the original topology on \(X\). Every \(d_X\)-open set is a union of such balls, so every \(d_X\)-open set is originally open.

Conversely, let \(U\) be originally open and \(x\in U\). Because \(e^{-1}:e(X)\to X\) is continuous, \(e(U)\) is open in the subspace \(e(X)\): indeed, \(e(U)=(e^{-1})^{-1}(U)\). By the definition of the subspace topology, there is a \(Q\)-open set \(V\) such that \(e(U)=V\cap e(X)\). Since \(e(x)\in V\) and the \(d_Q\)-metric induces the topology of \(Q\), there is \(r>0\) with \(B_{d_Q}(e(x),r)\subseteq V\). If \(y\in B_{d_X}(x,r)\), then \(e(y)\in B_{d_Q}(e(x),r)\subseteq V\), and \(e(y)\in e(X)\). Therefore \(e(y)\in V\cap e(X)=e(U)\), which implies \(y\in U\) by injectivity of \(e\). So \(B_{d_X}(x,r)\subseteq U\). Each point of \(U\) has a \(d_X\)-ball contained in \(U\), proving that \(U\) is \(d_X\)-open. The two topologies are equal. Applying the embedding theorem to a regular second-countable \(X\) proves the stated consequence. \(\square\)

The proof separates two requirements. Injectivity ensures that distinct points remain a positive distance apart. The embedding property, not just continuity, ensures that the transferred metric has enough open balls to recover the original topology.

A Common Pitfall: Continuity Is Not Enough

It is tempting to define \(d(x,y)=d_Q(f(x),f(y))\) for any continuous map \(f:X\to Q\) and immediately call \(d\) a metric. That conclusion can fail if \(f\) is not injective. For a concrete example, let \(X=\{a,b\}\) have the indiscrete topology \(\{\varnothing,X\}\), and let \(f:X\to Q\) be constant, with \(f(a)=f(b)=(0,0,\dots)\). The map is continuous, but the transferred formula gives \(d(a,b)=d_Q(f(a),f(b))=0\) even though \(a\ne b\). It is not a metric. The correction is to check injectivity before using the metric axioms; to recover the original topology, one also needs the inverse from the image to be continuous.

This transfer method matters because it converts a topological embedding into an explicit distance formula. In particular, once the embedding theorem supplies a map for a regular second-countable space, the pullback proves metrizability directly. It is not enough that points have coordinates: the coordinates must distinguish points and preserve the topology on the image.