Lebesgue's Number Lemma

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Theorem

Let $M = \left({A, d}\right)$ be a metric space.

Let $M$ be sequentially compact.


Then there exists a Lebesgue number for every open cover of $M$.


Proof

Aiming for a contradiction, suppose $\mathcal U$ is an open cover of $M$ for which no Lebesgue number exists.

Then for any $n \in \N$, there exists some $x_n \in M$ such that $B_{1/n} \left({x_n}\right) \subseteq U$ is false for each $U \in \mathcal U$. (Otherwise $1/n$ would be a Lebesgue number for $\mathcal U$.)

As $M$ is sequentially compact, $\left \langle {x_n} \right \rangle$ has a subsequence, say $\left \langle {x_{n \left({r}\right)}} \right \rangle$ which converges to some $x \in M$.

Since $\mathcal U$ covers $M$, $x \in U_0$ for some $U_0 \in \mathcal U$.

Since $U_0$ is open, $\exists m \in \N: B_{2/m} \left({x}\right) \subseteq U_0$.

Now $B_{1/m} \left({x}\right)$ contains $x_{n \left({r}\right)}$ for all $r \ge R$, say.

Choose $r \ge R$ such that $n \left({r}\right) \ge m$ and write $s = n \left({r}\right)$.

Then $B_{1/s} \left({x_s}\right) \subseteq B_{2/m} \left({x}\right)$ since:

\(\displaystyle d \left({x_s, y}\right)\) \(<\) \(\displaystyle \frac 1 s\)
\(\displaystyle \implies \ \ \) \(\displaystyle d \left({x, y}\right)\) \(\le\) \(\displaystyle d \left({x, x_s}\right) + d \left({x_s, y}\right)\)
\(\displaystyle \) \(<\) \(\displaystyle \frac 1 m + \frac 1 s\)
\(\displaystyle \) \(\le\) \(\displaystyle \frac 2 m\)

So $B_{1/s} \left({x_s}\right) \subseteq U_0$ which contradicts the choice of $x_s$.

So, by Proof by Contradiction, there has to be a Lebesgue number for $\mathcal U$.

$\blacksquare$


Source of Name

This entry was named for Henri Léon Lebesgue.


Sources