TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A sequentially compact metric space is totally bounded, proved from the axiom of dependent choice

Statement

Assume the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $\mathbb{N}$ -indexed chain). Let $(X,d)$ be a sequentially compact metric space (Countably compact, sequentially compact and limit point compact metric spaces, Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric). Then $(X,d)$ is totally bounded (Finite $\varepsilon$ -net and totally bounded metric space).

What is claimed about the cost, and what is not. Claimed: the proof below is carried out in $\mathrm{ZF} + \mathrm{DC}$ , and $\mathrm{DC}$ is used exactly once, at step 5.1. Not claimed: that $\mathrm{DC}$ is necessary for the statement. Establishing necessity would mean separating the statement from ZF, which is an independence result, and this library proves none. The reason countable choice is not used instead is that the point added at each stage has to be at distance at least $\varepsilon$ from the points already produced, so the set it is drawn from depends on the earlier stages; the first remark below spells that out.

Facts & Assumptions

Given: A sequentially compact metric space $(X,d)$ , and the Axiom of Dependent Choice.

[L1]

$(X,d)$ is sequentially compact: every sequence in $X$ has a subsequence converging to a point of $X$ , along a strictly increasing index map (Countably compact, sequentially compact and limit point compact metric spaces, Sequences of reals: bounded, eventually, frequently, tails, subsequences, Convergence of a sequence in a metric space: $x_k \to x$ iff $d(x_k, x) \to 0$ in $\mathbb{R}$ , A strictly increasing index map satisfies $n_k \ge k$ ).

[L2]

$(X,d)$ is totally bounded when for every real $\varepsilon > 0$ there is a finite $F \subseteq X$ , empty or listable, with $X = \bigcup_{y \in F} B(y,\varepsilon)$ ; equivalently, when for every real $\varepsilon > 0$ some finite list $y_0, \dots, y_{m-1}$ of points of $X$ satisfies: every $x \in X$ has $d(x,y_i) < \varepsilon$ for some $i < m$ (Finite $\varepsilon$ -net and totally bounded metric space, Open ball, closed ball and sphere in a metric space).

[L3]

Dependent choice: for a nonempty set $S$ , a relation $R$ on $S$ with every element $R$ -related to some element, and any $a \in S$ , there is a sequence $(t_n)$ in $S$ with $t_0 = a$ and $t_n \mathbin{R} t_{n+1}$ for every $n$ (The axiom of dependent choice: a relation in which every element is related to something admits an $\mathbb{N}$ -indexed chain).

[L4]

A convergent sequence is Cauchy: if $y_{n_j} \to p$ then for every rational $\varepsilon > 0$ there is $K$ with $d(y_{n_j}, y_{n_l}) < \varepsilon$ for all $j,l \ge K$ (Every convergent sequence in a metric space is Cauchy, Cauchy sequence in a metric space).

[L5]

A metric is symmetric, nonnegative and satisfies the triangle inequality (Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L6]

For every real $\eta > 0$ there is a natural $N \ge 1$ with $1/N < \eta$ , and $1/N$ is a positive rational (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , Every complete ordered field is Archimedean).

Proof

technique · contradiction

1.1

Suppose $(X,d)$ is sequentially compact and not totally bounded, and fix a real $\varepsilon > 0$ for which no finite subset of $X$ is an $\varepsilon$ -net.

L1L2assume-contra

2.1

Then $X \ne \emptyset$ , since for $X = \emptyset$ the empty set is an $\varepsilon$ -net.

L2step 1.1

3.1

Let $S$ be the set of $\varepsilon$ -separated finite tuples in $X$ , that is of functions $t : m \to X$ with $m \in \mathbb{N}$ and $d(t(i),t(j)) \ge \varepsilon$ whenever $i < j < m$ ; the empty function, with $m = 0$ , lies in $S$ , so $S \ne \emptyset$ .

L5step 2.1

4.1

Let $t \mathbin{R} t'$ mean that $t' : m+1 \to X$ extends $t : m \to X$ by one term with $d(t'(m), t(i)) \ge \varepsilon$ for every $i < m$ ; then $R$ is a relation on $S$ and every $t \in S$ is $R$ -related to some element of $S$ , because the finite set $\{t(0), \dots, t(m-1)\}$ is not an $\varepsilon$ -net, so some $x \in X$ has $d(x,t(i)) \ge \varepsilon$ for every $i < m$ , and the extension of $t$ by $x$ lies in $S$ .

L2L5step 3.1

5.1

Dependent choice, applied to $S$ , to $R$ and to the empty function as starting point, yields a sequence $(t_n)$ in $S$ with $t_0$ the empty function and $t_n \mathbin{R} t_{n+1}$ for every $n$ ; this is the only appeal to a choice principle in the proof.

L3step 4.1

6.1

Each $t_n$ has domain $n$ and $t_{n+1}$ restricted to $n$ is $t_n$ , both by induction on $n$ from the definition of $R$ ; so $y_n := t_{n+1}(n)$ defines a sequence $(y_n)$ in $X$ , and for $i < j$ both $y_i = t_{j+1}(i)$ and $y_j = t_{j+1}(j)$ hold, whence $d(y_i,y_j) \ge \varepsilon$ .

L5step 5.1

7.1

Sequential compactness gives a strictly increasing $j \mapsto n_j$ and $p \in X$ with $y_{n_j} \to p$ ; that subsequence is therefore Cauchy, so, taking a natural $N \ge 1$ with $1/N < \varepsilon$ and testing the Cauchy condition at the positive rational $1/N$ , there is $K \in \mathbb{N}$ with $d(y_{n_j}, y_{n_l}) < 1/N < \varepsilon$ for all $j,l \ge K$ .

L1L4L6step 6.1

8.1

But $n_K < n_{K+1}$ , so step 6.1 gives $d(y_{n_K}, y_{n_{K+1}}) \ge \varepsilon$ , contradicting step 7.1; the assumption of step 1.1 is therefore untenable, every real $\varepsilon > 0$ admits a finite $\varepsilon$ -net, and $(X,d)$ is totally bounded.

L2step 6.1step 7.1discharge-contradiction∎

Remarks

Why countable choice is not what this proof uses. A natural attempt is to apply $\mathrm{AC}_\omega$ to the family whose $n$ -th member is the set of $\varepsilon$ -separated $n$ -tuples, each of which is nonempty by the argument of step 4.1. What that returns is one $\varepsilon$ -separated $n$ -tuple for each $n$ , with no relation whatever between the tuple chosen at $n$ and the one chosen at $n+1$ : the tuples need not extend one another, need not share a single point, and nothing in the data assembles them into one $\varepsilon$ -separated sequence. The relation $R$ of step 4.1 is precisely the coherence that is missing, and building a sequence along a relation is what The axiom of dependent choice: a relation in which every element is related to something admits an $\mathbb{N}$ -indexed chain is. This is an observation about the argument given here; it is not a proof that $\mathrm{AC}_\omega$ is insufficient for the theorem.

The passage to a rational $\varepsilon$ in step 7.1. Convergence and the Cauchy condition are tested against rational $\varepsilon$ in this library (Convergence of a sequence in a metric space: $x_k \to x$ iff $d(x_k, x) \to 0$ in $\mathbb{R}$ , Cauchy sequence in a metric space), while the $\varepsilon$ of step 1.1 is an arbitrary positive real. The reciprocal form of the Archimedean property supplies a positive rational $1/N$ below it (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ ), and the contradiction is unaffected: a Cauchy estimate at $1/N \le \varepsilon$ still contradicts a separation of at least $\varepsilon$ .

This is the only implication on the page that costs dependent choice, and it is the reason For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice carries $\mathrm{DC}$ among its hypotheses. The full accounting is What each implication between the compactness properties of a metric space costs: which are theorems of ZF, which use countable choice, and which use dependent choice.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 90 results over 18 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Totally bounded space (Wikipedia) (standard reference, not scraped)
Axiom of dependent choice (Wikipedia) (standard reference, not scraped)
H. Herrlich, Axiom of Choice, Lecture Notes in Mathematics 1876, Springer 2006 (standard reference, not scraped)