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.
In a compact metric space,every open cover has a Lebesgue number.(Proof suggestions)
compactnessmetric-spacesproof-writingsolution-verification
Related Solutions
The statement of the problem is not so correct, since we are allowed to choose $p,q \in S$ such that $p=q$, viz. $d(p,q)=0$, so from here later we'll assume $p\neq q$. Let's go back to the problem now: since we assume the contraint $d(p,q)>r/2$ then $S$ is finite, i.e. there exists a integer $n\ge 1$ such that $|S|=n$ and $S:=\{p_1,p_2,\ldots,p_n\}$. It's well known that the propositions:
i) $X$ is totally limited
ii) $X$ is compact
iii) $X$ is sequentially compact
are equivalent (the easiest way to prove it is that (i) implies (ii) implies (iii) implies (i)).
So, there exists a finite collection of open balls $B(x_1,3r/4),B(x_2,3r/4),\ldots,B(x_{k_1},3r/4)$ that covers the whole compact metric space $(X,d)$. We can also assume without loss of generality that $B(x_i,3r/4) \cap X \neq \emptyset$ for all $1\le i\le k_1$.
Define $x_1:=\alpha_1$, and also $X_1:=X\setminus B(x_1,3r/4)$: if $x_1$ is empty then we ended (see below), otherwise $X_1\neq \emptyset$ is a metric space too, bounded, and closed, hence compact too. Then there exist a finite collection of open balls $B(y_1,3r/4),B(y_2,3r/4),\ldots,B(y_{k_2},3r/4)$ that covers $X_1$. Define $y_1:=\alpha_2$.
Repeat this algorithm infinitely many times, we have two cases:
1) If the sequence $\alpha_1,\alpha_2,\ldots$ is finite, then just define $p_i:=\alpha_i$ for all $i$ and we are done, indeed $d(p_i,p_j)\ge 3r/4 > r/2$ for all $1\le i < j \le n$.
2) If the sequence $\alpha_1,\alpha_2,\ldots$ is not finite, then the infinite collection of open balls $B(\alpha_1,3r/4),B(\alpha_2,3r/4),\ldots$ is a cover of $X$. Since $X$ is compact there exists a finite set of pairwise disjoint positive integers $T:=\{t_1,t_2,\ldots,t_n\}$ such that $B(\alpha_{t_1},3r/4),B(\alpha_{t_2},3r/4),\ldots,B(\alpha_{t_n},3r/4)$ is a cover too. Just set $\alpha_{t_i}=p_i$ for all $1\le i\le n$ and we really made our subcover of open balls that "do not overlap too much". []
First a notational observation: the notation $\{x\}^{(n)}$ is truly horrible. It looks as if you’re performing some operation on the singleton set $\{x\}$ whose sole element is $x$.
Your basic idea is just fine, but you’re making it too complicated; that’s part of why you’re having trouble expressing it clearly. As others have suggested, you don’t need to try to relate the centres of one ‘level’ to those of any other ‘level’. Here’s a relatively efficient version of the idea:
For each $n\in\Bbb Z^+$ let $$\mathscr{U}_n=\left\{B\left(x,\frac1n\right):x\in X\right\}\;;$$ $\mathscr{U}_n$ is an open cover of the compact space $X$, so there is a finite $F_n\subseteq X$ such that $$\left\{B\left(x,\frac1n\right):x\in F_n\right\}$$ covers $X$. Let $D=\bigcup_{n\in\Bbb Z^+}F_n$; $D$ is a countable union of finite sets, so $D$ is countable. To see that $D$ is dense in $X$ let $y\in X$ and $\epsilon>0$ be arbitrary. There is an $n\in\Bbb Z^+$ such that $\frac1n\le\epsilon$, and there is then an $x\in F_n$ such that $y\in B\left(x,\frac1n\right)$. But then $d(x,y)<\frac1n\le\epsilon$, so $y\in D\cap B(x,\epsilon)$, and $D$ is indeed dense in $X$. $\dashv$
You might like to note, by the way, that we could have got the same result had each of the sets $F_n$ been countable: we did not actually need them to be finite. Thus, the same argument shows that every Lindelöf metric space is separable. And this actually is a stronger result, since $\Bbb R$ with its usual metric is Lindelöf but not compact.
Best Answer
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?