TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01

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$L^{\infty} [0, 1]$ is not separable

Statement

The space $L^{\infty} ([0, 1])$ is not separable.

Facts & Assumptions

Given: The essential-supremum norm on $L^{\infty} ([0, 1])$ .

[L1]

Separability means having a countable dense subset (Separability: the existence of an at most countable dense subset, Finite, countably infinite, countable, uncountable).

[L2]

$L^{\infty}$ is normed by the essential supremum (The space $L^{\infty} (μ)$ of essentially bounded measurable functions, The $L^{p}$ norm descends to the quotient and makes $L^{p}$ a normed space for $1 \leq p \leq \infty$ ).

Proof

technique · direct

1.1L2givenalgebra

For each $t \in [0, 1]$ , let $u_{t} : = 1_{[0, t]}$ . If $s < t$ , then [L2, given, algebra] $u_{s}$ and $u_{t}$ differ by $1$ on $(s, t]$ , a set of positive measure, so $∥ u_{t} - u_{s} ∥_{\infty} = 1$ by [L2]. Thus the family ${u_{t} : t \in [0, 1]}$ is uncountable and $1$ -separated.

2.1L1step 1.1choosealgebra

Suppose $D$ were a countable dense subset. For each $t \in [0, 1]$ , choose [L1, step 1.1, choose, algebra] $d_{t} \in D$ with $∥ d_{t} - u_{t} ∥_{\infty} < 1 / 3$ . If $s \neq t$ , then the balls $B_{\infty} (u_{s}, 1 / 3)$ and $B_{\infty} (u_{t}, 1 / 3)$ are disjoint because the centers are distance $1$ apart, so $d_{s} \neq d_{t}$ . This gives an injection $[0, 1] ↪ D$ , contradicting countability.

3.1step 2.1∎

Therefore no countable subset is dense in $L^{\infty} ([0, 1])$ , so [step 2.1] $L^{\infty} ([0, 1])$ is not separable.

Depends on

Used by

The family {1_[0,t] : t ∈ [0,1]} is 1-separated in L^∞[0,1] Example

Dependency tree · two levels

17 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

Richard L. Wheeden and Antoni Zygmund, Measure and Integral: An Introduction to Real Analysis (standard reference, not scraped)

L∞[0,1]L∞[0,1] is not separable