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I am doing some extra exercises for an Analysis class, and I found this one. We haven’t seen much of what an open cover is, but I want to learn it. So, here it goes, and thank you everyone!

Let compact subset $S \in \mathbb{R} $, with $\mathcal O$ as its open cover. Complete to show that there exists some $\epsilon > 0$ so that for all $a \in S$, there is some $E \in \mathcal O$ such that $D(a, \epsilon) \subset E$.

a) Clearly state the negation of “There exists some $\epsilon > 0$ so that for all $a$ in $S$, there is some $E \in \mathcal O$ such that $D(a, \epsilon) \subset E$.”

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b) Assuming the above negation, explain why is true that: For each $n \in \mathbb{N}$, there is some $v_n \in S$ such that $D(v_n, 1/n)$ is not contained in any member of $\mathcal O$.

c) Since $S$ is compact, there is a subsequence $(v_{n_r})$ of $(v_n)$ that converges to some $v \in S$. Explain why there is some $G \in \mathcal O$ and some $\epsilon > 0$ such that $D(v, \epsilon ) \subset G$.

d) Explain why there is some $r \in \mathbb{N}$ such that that $1/n_r < \epsilon/2$ and

$|v_{n_r} – v| < \epsilon/2$.

e) Deduce: $D(v_{n_r} , 1/n_r) \subset D(v, \epsilon ) \subset G$.

f) Explain why we have reached a contradiction.

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You have the right idea. I will paraphrase what you wrote, so one gets a complete proof.

CLAIMLet $S\subseteq \bf R$ be compact. Let $\mathcal O$ be an open cover of $S$. Then there there exists an $\varepsilon >0$ such that for all $x\in S$ there is some $O\in \mathcal O$ such that $D(x,\varepsilon)\subseteq O$.

**PROOF** Suppose to the contrary. Then for every $\varepsilon’>0$ there is some $x\in S$ such that for no $O\in \mathcal O$ we have $D(x,\varepsilon’)\subseteq O$. Now for each $n=1,2,3,\ldots$ pick $\varepsilon_n=\dfrac 1 n$. Then we obtain a sequence $\langle v_1,v_2,v_3,\ldots\rangle $ in $S$ such that for no $O\in\mathcal O$ we have $D(v_n,n^{-1})\subseteq O$. Since $S$ is compact, there is $v\in S$ and a subsequence $\langle v_{n_1},v_{n_2},\ldots \rangle$ with $n_1<n_2<n_3<\cdots$ such that $\lim\limits_{k\to\infty} v_{n_k}=v$. Now, $\mathcal O$ covers $S$; so there exists $O’\in\mathcal O$ such that $v\in O’$. Since $O’$ is open, there exists $\varepsilon >0$ such that $D(v,\varepsilon)\subseteq O’$. Since $v_{n_k}\to v$, there exists $K$ such that for any $k\geqslant K$ we have $|v-v_{n_k}|<\dfrac \varepsilon 2$. If $n_K^{-1}<\dfrac \varepsilon 2$, we’re done, we pick $n_K$ and $v_K$. Else, we can pick $K_1>K$ large enough so that $n_{K_1}^{-1}<\dfrac \varepsilon 2$, and by the above we still have $|v-v_{n_{K_1}}|<\dfrac \varepsilon 2$. In any case, we have found what we wanted with $K=k$ or $=K_1$. Now, pick $x\in D(v_{n_k},n_k^{-1})$. Then $$|v-x|\leqslant |v-v_{n_k}|+|v_{n_k}-x|<\frac\varepsilon 2+\frac\varepsilon 2=\varepsilon$$

that is, $|v-x|<\dfrac \varepsilon 2$. It follows that $D(v_{n_k},n_k^{-1})\subseteq D(v,\varepsilon)\subseteq O’$. But this contradicts the fact that *no* ball $D(v_n,n^{-1})$ could be contained in an $O\in \mathcal O$. Thus our assumption that no such $\varepsilon$ existed must have been false, and the theorem is proven. $\blacktriangle$

**NOTE** The above proof applies in more generality to any metric space $(X,d)$ having the Bolzano-Weiertrass property, that is, the property that every infinite subset has an accumulation point in $X$. The proof is conceptually the same, but a point has to be taken care of: if the sequence obtained $\langle v_1,v_2,\ldots\rangle$ is of finite range, that is, the set $\{v_1,v_2,\ldots\}$ is finite, then some element, call it $v$, repeats infinitely often. Pick $O$ in the cover such that $v\in O$. Since $O$ is open, we obtain $\delta >0$ such that $B(v,\delta)\subseteq O$. If $v=v_{k}$, pick $k’>k$ such that $k’^{-1}<\delta$ and $v_{k’}=v$. This can be done because by assumption $v$ repeats infinitely often. Then $B(v_{k’},k’^{-1})\subseteq B(v,\delta)\subseteq O$, a contradiction. If not, the set is infinite and we obtain an accumulation point. Then the proof is exactly the same as the above, save we don’t talk about *subsequences*, rather work with the set itself, which has an accumulation point.

As a final comment, the number so obtained is called a Lebesgue number for the cover $\mathcal O$. Let $$L(\mathcal O)=\{\varepsilon >0:\varepsilon \text{ is a Lebesgue number for }\mathcal O\}$$ Then $\varepsilon_L=\sup L(\mathcal O)$ is called *the* Lebesgue number of $\mathcal O$. Thus, we have proven

PROPLet $(X,d)$ be a metric space with the Bolzano Weiertrass property. Then every open cover $\mathcal O$ of $(X,d)$ has a Lebesgue number.

In fact, with a little more work, one can prove

THMLet $(X,d)$ be a metric space. Then $X$ is compact $\iff$ it has the Bolzano Weiertrass property.

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