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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 ( ${A C}_{ω}$ )) and the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain). The family $D = {d_{a} : a \in [0, 1]} \subseteq C ([0, 1], R)$ , 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 $∣ d_{a} (x) - d_{b} (x) ∣ \leq ∣ a - b ∣$ for every $x$ ; evaluating at $x = 0$ gives $∥ d_{a} - d_{b} ∥_{\infty} = ∣ a - b ∣$ .

givenalgebra

2.1

Given a sequence $d_{a_{n}}$ in $D$ , 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 $D$ is sequentially compact, and it is compact by [L2].

step 2.1L2algebra∎

Depends on

Used by

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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↦∣x−a∣, a∈[0,1]a∈[0,1], is compact in C([0,1])C([0,1])