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Under the Axiom of Countable Choice and the Axiom of Dependent Choice, the family $x\mapsto|x-a|$ , $a\in[0,1]$ , is compact in $C([0,1])$

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( $\mathrm{AC}_\omega$ )) and 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). The family $\mathcal D=\{d_a:a\in[0,1]\}\subseteq C([0,1],\mathbb R)$ , where $d_a(x)=|x-a|$ , is compact in the supremum metric.

Facts & Assumptions

Given: The Axiom of Countable Choice, the Axiom of Dependent Choice, and $d_a(x)=|x-a|$ for $a,x\in[0,1]$ .

[L1]

Every sequence in $[0,1]$ has a convergent subsequence (Bolzano-Weierstrass: every bounded real sequence has a convergent subsequence).

[L2]

Assuming the Axiom of Countable Choice and the Axiom of Dependent Choice, sequential compactness and compactness are equivalent for a metric space (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).

Proof

technique · sequential

1.1

The reverse triangle inequality gives $|d_a(x)-d_b(x)|\le|a-b|$ for every $x$ ; evaluating at $x=0$ gives $\lVert d_a-d_b\rVert_\infty=|a-b|$ .

givenalgebra

2.1

Given a sequence $d_{a_n}$ in $\mathcal D$ , use [L1] to choose $a_{n_j}\to a\in[0,1]$ . Step 1.1 then gives $d_{a_{n_j}}\to d_a$ uniformly.

L1step 1.1choose

3.1

Thus $\mathcal D$ is sequentially compact, and it is compact by [L2].

step 2.1L2algebra∎

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Sources

The Ascoli--Arzelà Theorem (MIT) (standard reference, not scraped)

Under the Axiom of Countable Choice and the Axiom of Dependent Choice, the family x↦∣x−a∣x\mapsto|x-a|x↦∣x−a∣, a∈[0,1]a\in[0,1]a∈[0,1], is compact in C([0,1])C([0,1])C([0,1])