Exercise · ZW-LEBESGUE-NUMBER-010 · ★★★ Advanced

\(\mathbb{R}\) is not compact. Can an open cover of \(\mathbb{R}\) have a Lebesgue number?

Choose one option.

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  • A. Yes: the cover \(\{(n-1,n+1)\}_{n\in\mathbb{Z}}\) admits \(\delta=1\)

Explanation

A set of diameter less than \(1\) with infimum \(a\) fits in \((n-1,n+1)\) taking \(n=\lceil a\rceil\) (if \(a\) is an integer, \(n=a\)). Compactness guarantees a Lebesgue number for every cover, but a particular cover can have one without it.

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Review the concept: Lebesgue number

A Lebesgue number of an open cover \(\mathcal{U}\) of a metric space is a \(\delta>0\) such that every subset \(A\) of diameter less than \(\delta\) fits inside some member of \(\mathcal{U}\).

\[\operatorname{diam}A<\delta\ \Rightarrow\ \exists\,U\in\mathcal{U}:\ A\subseteq U\]

More practice problems: lebesgue number

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