topology of 𝒞

Denote the Cantor space $𝒞=\{0,1\}^ℕ$.

Definition

For each $i∈ℕ$, denote the $i$-th projection map $π_i:𝒞→\{0,1\}$, $π_i(x)=x(i)$.

Definition

The product topology on $𝒞$ is the smallest topology such that the projection maps are continuous. In particular, since $\{0,1\}$ is discrete, we must have that

• $π_i^{-1}(\{0\}) = \{x∈C:x(i)=0\}$
• $π_i^{-1}(\{1\}) = \{x∈C:x(i)=1\}$

(i.e., sets determined by a single index) are open sets for all $i∈ℕ$. This implies also that all finitely-determined sets are open.

Definition

define the normalized Hamming metric on $𝒞$ by

$$d(x,y) = ∑_{i=0}^∞ \frac {|xᵢ-yᵢ|} {2ⁱ⁺¹}.$$

Problem

Is the product topology equivalent to the topology induced by the (normalized) Hamming metric?

Lemma

$U$ is open in the product topology iff $U$ is a union of finitely-determined sets.

Proof

Define $𝒯$ to be collection of all unions of finitely-determined sets. Observe that $𝒯$ is a topology:

• the empty union gives the empty set.
• the null-determined set (set with no constraints) is all of $𝒞$.
• unions of unions are unions.

Finite intersections: let $U=⋃_{A∈𝒜} A$ and $V=⋃_{B∈ℬ} B$, where $𝒜,ℬ$ are collections of f.d. sets. Then

$$U∩V = ⋃_{A∈𝒜,B∈ℬ} A∩B,$$

and notice that each $A∩B$ is still f.d.

Also note that $𝒯$ contains all sets of the form $ϕᵢ⁻¹(b)$ since each of these is f.d. So the product topology is a subset of $𝒯$ (by minimality).

So if $U$ open in the product topology, then $U$ open in $𝒯$.

Conversely suppose $U∈𝒯$. Say $U=⋃_{A∈𝒜} A$ with each $A$ f.d. Then each $A$ is a union of (finite intersections of $ϕᵢ⁻¹(\{0\})$s or $ϕᵢ⁻¹(\{1\})$s) and so is open; and $U$ is an arbitrary union of open sets, which is also open.

Claim

$U$ is open in the product topology iff it is open under the (normalized) Hamming metric.

Proof

• [⟹] Suppose $U$ is open in the product topology. Then $U=⋃_{A∈𝒜} A$ for some collection of f.d. sets $𝒜$.

  We want to show that $U$ is open under the NH metric; i.e., for all $x∈U$ there exists $ϵ>0$ so that $B_ϵ(x)⊆U$.

  Fix $x∈U$. Then $x∈A$ for some f.d. set $A∈𝒜$. Since $A$ is finitely-determined, let $k$ be the last index determining membership in $A$. That is, indices above $k$ don’t matter—for all $x,y$ agreeing on the first $k$ coordinates, $x,y$ are either both in $A$ or both not in $A$.

  Let $ϵ=1/2ᵏ⁺¹$. Suppose $y∈B_ϵ(x)$. Notice that if $xᵢ≠yᵢ$ then $d(x,y)≥1/2ⁱ⁺¹$. So $d(x,y)<1/2ᵏ$ ensures that $x$ and $y$ agree on (at least) the first $k$ digits. So then the fact that $x∈A$ implies $y∈A$ as well, and $A⊆U$ implies $y∈U$ also. Thus $B_ϵ(x)⊆U$.

• [⟸] Conversely, suppose $U$ is open under NH metric.

  We want to show that $U$ is open under product topology.

  Fix $x∈U$, and by metric-openness there exists $ϵ>0$ such that $B_ϵ(x)⊆U$. Let $k∈ℕ$ so that $1/2ᵏ≤ϵ$. Let $B'$ be the set of all strings agreeing with $x$ on (at least) the first $k$ digits. Then if $y∈B'$,

  $$d(x,y) ≤ ∑_{i=k+1}^∞ \frac 1 {2^{i+1}} ≤ \frac 1 {2^{k+1}} < ϵ \qquad⟹\qquad y∈B_ϵ(x).$$

  Thus $B'⊆B_ϵ(x)$. Also note that $B'$ is f.d., so $B'$ is open in the product topology.

  In particular we just established that for each $x∈U$, there exists a $B'(x)⊆U$ such $B'(x)$ is open in the product topology. Then $U=⋂_{x∈U} B'(x)$, and furthermore this establishes that $U$ is open in the product topology.