Lebesgue number

Let $(𝑋, 𝑑)$ be a compact metric space and $U$ be an open cover. Then there exists a Lebesgue number $𝜆 > 0$ such that every $𝑌 \subseteq 𝑋$ with diameter less than $𝜆$ is contained entirely within one of the covering sets, #m/thm/anal i.e.

d i a m (𝑌) < 𝜆 ⟹ (\exists 𝑈 \in U) [𝑌 \subseteq 𝑈]

Proof

Since $U$ is an open cover, for every $𝑥 \in 𝑋$ there exists some neighbourhood $𝑈 \in U$ of $𝑥$ , and hence some $𝛿 𝑥 > 0$ so that $B 𝛿 𝑥 (𝑥) \subseteq 𝑈$ . Then ${B 𝛿 𝑥 (𝑥) : 𝑥 \in 𝑋}$ is an open cover, and since $𝑋$ is compact there exists some finite subcover ${B 𝛿 𝑥 𝑖 (𝑥 𝑖)} 𝑛 𝑖 = 1$ where $(𝑥 𝑖) 𝑛 𝑖 = 1$ are points in $𝑋$ . Then $𝜆 = m i n {𝛿 𝑥 𝑖} 𝑛 𝑖 = 1$ is a Lebesgue number.

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