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Arzelà--Ascoli for real $C(K)$ under Countable Choice and Dependent Choice: compact closure iff equicontinuous and pointwise bounded

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). Let $K$ be a nonempty compact metric space and $\mathcal F\subseteq C(K,\mathbb R)$ . Its closure in the supremum metric is compact if and only if $\mathcal F$ is equicontinuous and pointwise bounded.

Facts & Assumptions

Given: The Axiom of Countable Choice, the Axiom of Dependent Choice, and a family $\mathcal F\subseteq C(K,\mathbb R)$ .

[L1]

An equicontinuous pointwise-bounded family is totally bounded in the supremum metric (An equicontinuous pointwise-bounded family in $C(K,\mathbb R)$ has a finite net in the supremum metric).

[L2]

$C(K,\mathbb R)$ is complete in the supremum metric ( $C(K,\mathbb{R})$ is complete in the supremum metric for every nonempty compact metric space $K$ ).

[L3]

A subspace of a complete metric space is complete exactly when it is closed; assuming Countable Choice and Dependent Choice, in a metric space compactness is equivalent to completeness together with total boundedness (A subspace of a complete metric space is complete iff it is closed, and a complete subspace of any metric space is closed, 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).

[L4]

Equicontinuity and pointwise boundedness are as defined in Equicontinuity, pointwise boundedness, and uniform boundedness for families in $C(K,\mathbb R)$ .

[L5]

A continuous function on a compact metric space is uniformly continuous (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous).

Proof

technique · direct

1.1

Suppose $\mathcal F$ is equicontinuous and pointwise bounded. By [L1] it is totally bounded, and its closure is totally bounded as well.

L1algebra

1.2

Conversely suppose the closure is compact. For a positive $\varepsilon$ , choose a finite $\varepsilon/3$ -net $g_0,\ldots,g_N$ in the closure; by [L5], a common positive radius makes every $g_i$ vary by less than $\varepsilon/3$ .

L3L5choose

2.1

The closure is closed in the complete space of [L2], hence complete by [L3]. Therefore its closure is compact by [L3].

step 1.1L2L3

2.2

For $f\in\mathcal F$ , choose $g_i$ within $\varepsilon/3$ in supremum distance. The two uniform-distance bounds and step 1.2 give $|f(x)-f(y)|<\varepsilon$ whenever $d(x,y)$ is below the common radius.

step 1.2algebra

2.3

The same finite net bounds $|f(a)|$ at each fixed $a\in K$ , so $\mathcal F$ is pointwise bounded.

step 1.2L4algebra

3.1

Steps 2.2 and 2.3 give equicontinuity and pointwise boundedness, completing the converse.

step 2.2step 2.3L4∎

Depends on

Used by

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Sources

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

Arzelà--Ascoli for real C(K)C(K)C(K) under Countable Choice and Dependent Choice: compact closure iff equicontinuous and pointwise bounded