Open Cover for a Compact Subset

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 $D(a, \epsilon) \subset E$.”

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.

Solutions Collecting From Web of "Open Cover for a Compact Subset"

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

CLAIM Let $S\subseteq \bf R$ 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

PROP Let $(X,d)$ 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

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