In a compact metric space,every open cover has a Lebesgue number.(Proof suggestions)

Question

I started the proof as follows.Suppose $X$ is compact.We want to show that every open cover of $X$ has a Lebesgue number.We first show that every open ball cover of $X$ has a Lebesgue number.Take an open ball cover $\{B_\alpha:\alpha \in I\}$,then by compactness there is a finite subcover that is there is a collection of balls $B(x_1,r_1),...,B(x_k,r_k)$ covering $X$. Assume that no ball is contained in other(to remove redundacy).Automatically it is implied that the centres are distinct(if not ,then larger ball contains smaller one,not possible as per our assumption).Now,suppose $\delta=1/2 \min\{r_n:n=1,2,..,k\}=r/2>0$.Now I am not sure whether this will be a Lebesgue number of the finite subcover.Can someone please tell me what to do next or whether I am proceeding in a right way.

Answer 1

That's not really the way to go.

A contradiction proof is better. Assuming that $\mathcal U$ is an open cover of $X$ that has no Lebesgue number, it follows in particular for each integer $n > 1$ the fraction $\lambda = \frac{1}{n}$ is not a Lebesgue number for $\mathcal U$, and therefore the exists a subset $A_n \subset X$ such that the diameter of $A_n$ is less than $\frac{1}{n}$, but $A_n$ is not contained in any element of the given cover.

The idea now is to choose a sequence $(x_n)$ in $X$ such that $x_n \in A_n$ for all $n$. Some subsequence $(x_{n_i})$ converges to a limit $p \in X$. This point $p$ is contained in some $U \in \mathcal U$, and so there exists $r > 0$ such that $B(p,r) \subset U$.

Perhaps you can now see how derive a contradiction, by taking the subsequence index $i$ to be sufficiently large?

Answer 2

There is indeed a direct proof (I know) of the Lebesgue number lemma. Take an open cover $\mathcal U$ of $X.$ For any $x \in X$ there exists $U \in \mathcal U$ and an $r_x \gt 0$ such that $B (x, r_x) \subseteq U.$ Now consider another open cover $$\mathcal U' : = \left \{B \left (x, \frac {r_x} {2} \right )\ \bigg |\ x \in X \right \}$$ of $X.$ Since $X$ is compact $\mathcal U'$ has a finite subcover say $$\mathcal U_F' : = \left \{B \left (x_i, \frac {r_{x_i}} {2} \right )\ \bigg |\ 1 \leq i \leq n \right \}$$ for some $n \in \mathbb N.$ Let $r : = \min \left \{\frac {r_{x_i}} {2}\ \bigg |\ 1 \leq i \leq n \right \} \gt 0.$

Claim $:$ $r$ is the Lebesgue number for the open cover $\mathcal U$ of $X.$

To show that take any $y \in X$ arbitrarily. We need to show that $B (y,r)$ is contained in some open set in the open cover $\mathcal U.$ Since $\mathcal U_{F}'$ covers $X$ there exists $1 \leq j \leq n$ such that $y \in B \left (x_j, \frac {r_{x_j}} {2} \right ).$ This implies that $d (y, x_j) \lt \frac {r_{x_j}} {2},$ where $d$ is the metric endowed with $X.$ Now let $z \in B (y, r)$ and let us try to estimate the distance between $z$ and $x_j$ with the help of triangle inequality. The idea here is to show that $z \in B \left (x_j, r_{x_j} \right )$ and since $B \left (x_j, r_{x_j} \right)$ is contained in some open set in the open cover $\mathcal U$ we are done with the proof of the claim. Now,

$$\begin{align*}d \left (z, x_j \right ) & \leq d (z, y) + d \left (y, x_j \right ) \\ & \lt r + \frac {r_{x_j}} {2} \\ & \leq \frac {r_{x_j}} {2} + \frac {r_{x_j}} {2} \\ & = r_{x_j}. \end{align*}$$

Therefore, $z \in B \left (x_j, r_{x_j} \right ),$ as required.

This proves the claim and the proof of the theorem is thus completed as well.