ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-09-01

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The family ${1_{[0, t]} : t \in [0, 1]}$ is $1$ -separated in $L^{\infty} [0, 1]$

Example

The family ${1_{[0, t]} : t \in [0, 1]}$ is $1$ -separated in $L^{\infty} ([0, 1])$ .

Facts & Assumptions

Given: Two parameters $s, t \in [0, 1]$ with $s \neq t$ .

[L1]

$L^{\infty} ([0, 1])$ is not separable, and the proof runs through this explicit family ( $L^{\infty} [0, 1]$ is not separable).

Verification

technique · direct

1.1L1givenalgebra

Assume $s < t$ . Then [L1, given, algebra] $1_{[0, t]} - 1_{[0, s]}$ equals $1$ on $(s, t]$ , a set of positive measure.

2.1step 1.1algebra

Hence [step 1.1, algebra] $∥ 1_{[0, t]} - 1_{[0, s]} ∥_{\infty} = 1.$ The same conclusion holds when $t < s$ by symmetry.

3.1L1step 2.1∎

So any two distinct members of the family are distance $1$ apart, which is [L1, step 2.1] exactly the one-separated property used in [L1].

Depends on

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

Used by

Nothing in the library uses this result yet.

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

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

The family {1[0,t]:t∈[0,1]}{1[0,t]​:t∈[0,1]} is 11-separated in L∞[0,1]L∞[0,1]