Under Dependent Choice, every nonempty compact metric space is a continuous image of Cantor space

Given: The objects, hypotheses, and choice principles stated above.

If $K$ is a nonempty compact metric space, there is a rooted levelled tree $T={\bigcup }_{n\in \mathbb{N}}{T}_n$, finitely branching with every level $T_n$ finite and nonempty — so $T$ itself is infinite — and nonempty compact sets $(K_s{)}_{s\in T}$ such that $T_0$ has one root $r$ with $K_r=K$, every node has a finite nonempty set of children whose sets cover its set, every child set is contained in its parent set, and $\mathrm{diam}(K_s)\le 2^{-n}$ for $s\in T_n$ after a harmless rescaling of the metric. (Under Dependent Choice, a compact metric space carries a finitely branching refining tree of covers of arbitrarily small diameter).

Let $(X,d)$ be a compact metric space (def-metric-compactness, def-metric-space). Then $(X,d)$ is totally bounded (def-totally-bounded) and complete (def-complete-metric-space). Both implications are theorems of ZF. Completeness is obtained here from the finite intersection characterisation (thm-compact-iff-finite-intersection-property) applied to the closures of the tails of a Cauchy sequence, and not from the extraction of a convergent subsequence, which would route the argument through sequential compactness. What matters for the ledger is that the route taken below selects nothing at all; the first remark below says why the other route was not taken. (A compact metric space is complete and totally bounded, and neither implication uses any choice principle).

Under Countable Choice for assertion 1, let $(X,d)$ be a metric space (def-metric-space). Call a sequence $(F_k{)}_{k\in \mathbb{N}}$ of subsets of $X$ a Cantor chain if every $F_k$ is nonempty, closed (def-metric-topology) and bounded, $F_{k+1}\subseteq F_k$ for every $k$, and $\mathrm{diam}(F_k)\to 0$ in $\mathbb{R}$ (def-metric-bounded-diameter, def-real-limit). Then: 1. If $(X,d)$ is complete (def-complete-metric-space), every Cantor chain in $X$ has an intersection ${\bigcap }_{k\in \mathbb{N}}{F}_k$ with exactly one element. 2. Conversely, if every Cantor chain in $X$ has nonempty intersection, then $(X,d)$ is complete. Boundedness of each $F_k$ is part of the definition of a Cantor chain because $\mathrm{diam}$ is defined for nonempty bounded sets only in this library (def-metric-bounded-diameter); it is not an extra hypothesis but the precondition for writing the diameter condition down. (In a complete metric space nested nonempty closed sets whose diameters tend to $0$ meet in exactly one point, and this property characterises completeness).

The product set. Let $I$ be a set and let $X_i$ be a set for each $i\in I$. The product is ${\prod }_{i\in I}{X}_i:=\{x:x\text{ is a function with domain }I\text{ and }x(i)\in X_i\text{ for every }i\in I\},$ and we write $x_i:=x(i)$, the $i$-th coordinate of $x$. Two elements of the product are equal exactly when they agree at every index, functions being equal when they have the same domain and the same values. For $j\in I$ the $j$-th projection is ${\pi }_j:{\prod }_{i\in I}{X}_i\to X_j,{\pi }_j(x):=x_j\mathrm{.}$. The product topology ${\mathcal{T}}^{\mathrm{\Pi }}$ on ${\prod }_i{X}_i$ is the initial topology of the projections: the topology generated by the subbasis $\{{\pi }_i^{-1}[U]:i\in I,U\in {\mathcal{T}}_i\}$. Finite intersections of subbasic sets form a basis for it, and they are exactly the boxes ${\prod }_{i\in I}{U}_i$ with every $U_i$ open in $X_i$ and $U_i=X_i$ for all but finitely many $i$. (The product set ${\prod }_{i\in I}{X}_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space).

Throughout, a topology is as in def-topological-space, and finite, at most countable and uncountable are as in def-countable, so that "countable" always means "at most countable" and every finite set is countable. Let $X$ be a set. The six families below are topologies on $X$; that each really satisfies (T1), (T2) and (T3) is discharged in full after the list. Among those six is the discrete topology ${\mathcal{T}}_{\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{c}}:=\mathcal{P}(X)$, in which every subset is open and hence every subset is also closed. (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

Let $X$ be a set and let $R\subseteq X\times X$ be a binary relation on $X$. Call $R$ entire on $X$ when $\text{for every }x\in X\text{ there is }y\in X\text{ with }xRy\mathrm{.}$ The Axiom of Dependent Choice, written $\mathrm{D}\mathrm{C}$, is the following statement. The statement is: for every nonempty set $X$, every relation $R$ entire on $X$, and every $a\in X$, there is a sequence $x:\mathbb{N}\to X$ with $x_0=a$ and $x_{n}R{x}_{n+1}$ for every $n\in \mathbb{N}$. (The axiom of dependent choice: a relation in which every element is related to something admits an $\mathbb{N}$-indexed chain).

Dependent Choice implies Countable Choice directly: for a sequence $(A_n{)}_{n\in \mathbb{N}}$ of nonempty sets, let $S$ be the set of finite sequences $s$ with $s(i)\in A_i$ for $i<\mathrm{length}(s)$. The empty sequence belongs to $S$. Relate $s$ to each extension by one entry from $A_{\mathrm{length}(s)}$; this relation is entire because that set is nonempty. Apply [F6] starting at the empty sequence. The resulting nested sequences have lengths $0,1,2,\dots$, and their union is a function choosing an element of every $A_n$. No simultaneous choices were used to establish that the relation is entire.