I want to prove
Lebesgue's number lemma: If the metric space $(X,d)$ is compact and an open cover of $X$ is given, then there exists a number $\delta>0$ such that for every $x\in X$, the open ball $B(x;\delta)$ is contained in some member of the cover.
My attempt: If $X=\varnothing$, then the lemma is clearly true so we may assume that $X\neq\varnothing$. Let $U_1,\ldots,U_n$ a finite open cover of $X$ and set $U=U_1\cup\ldots \cup U_n$. First, Suppose that $\partial U\neq\varnothing$ (where $\partial U$ denotes the boundary of $U$) and consider the set
$$
D=\{d(x,y):x\in X,y\in \partial U\}.
$$
Since $\partial U\neq\varnothing$ and $X\neq\varnothing$, it follows that $D\neq\varnothing$, so $\delta=\mathrm{inf}(D)$ exists. I claim that $\delta>0$. Suppose otherwise that $\delta=0$. Therefore, for every positive integer $n$, there exist $x_n\in X$, $y_n\in\partial U$ such that $d(x_n,y_n)<1/n$. Since the sequence $(x_1,x_2,x_3,\ldots)$ is in $X$ and $X$ is compact, there exist a convergence sub-sequence $(x_{n_1},x_{n_2},x_{n_3},\ldots)$. Let $x_0\in X$ be its limit.
Now, given any $\varepsilon>0$, choose $k_1$ such that
$d(x_{n_k},x_0)<\varepsilon/2$ for every $k_1\le k$. In addition, choose $k_2$ such that $1/n_{k_2}<\varepsilon/2$. Hence, if $k_0=\max\{k_1,k_2\}$, then
$$
d(y_{n_k},x_0)\le d(y_{n_k},x_{n_k})+d(x_{n_k},x_0)
<\frac{1}{n_k}+\frac{\varepsilon}{2}<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon
$$
for every $k_0\le k$. Therefore, $(y_{n_1},y_{n_2},y_{n_3},\ldots)$ converges to $x_0$ also. But this sequence is in $\partial U$ and since $\partial U$ is closed, it follows that $x_0\in \partial U$. On the other hand, $U\cap \partial U=\varnothing$ since $U$ is open, so $x_0\notin U$. But that is impossible since $U$ covers $X$. Thus $\delta>0$.
Now, given $x\in X$, there exist $\epsilon>0$ such that $B(x;\epsilon)\subseteq U_i$ for some $i$. But $\epsilon\le \delta$, so the ball $B(x;\delta)$ is contained in $U_i$, as required.
My questions are:
(a) is this proof correct?
(b) How to prove the lemma in the case where $\partial U=\varnothing$ ?
Thanks!
Best Answer
This proof cannot work. Note that $\delta=0$ trivially whenever $\partial U \neq \emptyset$. Moreover it doesn't matter if we start with a finite cover or not. Even if we just have two clopen sets in the cover (e.g. as in the cover $(\leftarrow, \sqrt{2}), (\sqrt{2}, \rightarrow)$ of $\Bbb Q$), for any $\delta>0$ we have points in these two cover sets that are $< \delta$ apart, so this $\delta$ could not work as a Lebsgue number. We essentially need the compactness and so for any cover $\mathcal{U}$ of $X$ we can pick $U_x \in \mathcal{U}$ and $r_x >0$ so that $B(x, 2r_x) \subseteq U_x$. Then take a finite subcover of the new open cover $\{B(x, r_x)\mid x \in X\}$ and apply compactness of $X$ to that to finish the proof in tehe right way.
Staying away from the boundary for the original cover won't work; it's not enough.