ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-09-01

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$c_{0}$ is Banach for the supremum norm

Example

Let $c_{0} : = {x = (x_{n})_{n \geq 0} \in ℓ^{\infty} : x_{n} \to 0} .$ With the supremum norm inherited from $ℓ^{\infty}$ , the space $c_{0}$ is Banach.

Facts & Assumptions

Given: A sequence $x = (x_{n})$ in $ℓ^{\infty}$ that is a supremum-norm limit of a sequence $(x^{(m)})$ in $c_{0}$ .

[L1]

The space $ℓ^{\infty}$ is Banach for the supremum norm ( $ℓ^{\infty}$ is Banach for the supremum norm).

[L2]

A closed subspace of a Banach space is Banach (A closed subspace of a Banach space is Banach).

Verification

technique · direct

1.1

Fix $ε > 0$ and choose $m$ with $∥ x - x^{(m)} ∥_{\infty} < ε / 2$ . Because $x^{(m)} \in c_{0}$ , there is $N$ with $∣ x_{n}^{(m)} ∣ < ε / 2$ for all $n \geq N$ .

givenchoose

2.1

For $n \geq N$ , $∣ x_{n} ∣ \leq ∣ x_{n} - x_{n}^{(m)} ∣ + ∣ x_{n}^{(m)} ∣ < ε / 2 + ε / 2 = ε$ . Hence $x_{n} \to 0$ , so $x \in c_{0}$ .

step 1.1algebra

3.1

Step 2.1 shows that $c_{0}$ is closed in $ℓ^{\infty}$ . Since $ℓ^{\infty}$ is Banach by [L1], [L2] makes $c_{0}$ Banach.

step 2.1L1L2∎

Depends on

Used by

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Theo Buhler and Dietmar A. Salamon, Functional Analysis (standard reference, not scraped)
Andrew Lin and Casey Rodriguez, MIT 18.102 Introduction to Functional Analysis (standard reference, not scraped)

c0c0​ is Banach for the supremum norm