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I am absolutely stuck on this analysis problem from Taylor's Foundations of Analysis:

"Prove that if $I$ is a closed, bounded interval which is contained in the union of some collection of open intervals, then $I$ is contained in the union of some finite subcollection of these open intervals."

I have just about no ideas on where even to start; I've tried direct proof and got nowhere and contradiction by supposing that any finite subcollection will leave out points of $I$. I was trying to make the argument that there would have to be infinitely many such points, and thus that a finite subcollection could not contain such a set, but realised that I'm pretty sure that only works if the set of points is discrete, which I can't guarantee. So could anyone give a suggestion on where to start (or where to go if I'm actually starting off correctly)?

If it helps, this in the chapter on sequences, in the section on Cauchy sequences.

amWhy
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themathandlanguagetutor
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  • Heine-Borel Theorem states that every closed and bounded subset of $\Bbb R^n$ is compact. We say that a set is compact if every open cover has a finite subcover. However, there are also other ways to prove compactness of $[a,b]$. See this post: https://math.stackexchange.com/questions/368108/how-to-prove-every-closed-interval-in-r-is-compact – ThePortakal Jul 17 '17 at 22:11
  • I wonder if there is a way around this though, since this particular text doesn't make any mention of compactness and doesn't introduce the Heine-Borel theorem. I will look into it nonetheless – themathandlanguagetutor Jul 17 '17 at 22:14

2 Answers2

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If $I=[a,b]$, let $S$ be the set$$\{x\in[a,b]\,|\,[a,x]\text{ is contained in the union of a finite subcollection of the open intervals}\}.$$You want to prove that $b\in S$. Well, $a\in S$; that's trivial. In particular, $S\neq\emptyset$. Let $s=\sup S$. It is not hard to get a contradiction if $s<b$. Thes $s=b$. Now use the fact that $b$ belongs to one of the intervals to deduce that $b\in S$.

José Carlos Santos
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Here is another standard approach.

Set $a_0 = a$ and $b_0 = b$.

Suppose that $[a,b]$ cannot be covered by finitely many of the open intervals. Then the same must be true of at least one of the subintervals $[a, (a+b)/2]$ or $[(a+b)/2, b]$. Let $[a_1, b_1]$ denote one of these subintervals which can't be covered by finitely many open intervals.

Continue this subdivision repeatedly. At the $n$'th stage we start with an interval $[a_n, b_n]$ of length $(b-a)/2^n$ which can't be covered by finitely many of the open intervals, and we produce a subinterval $[a_{n+1}, b_{n+1}]$ of half the length, with the same property.

Observe that $a_0 \leq a_1 \leq \cdots$ and for each $n$, we have $a \leq a_n \leq b$. This means that $\{a_n\}$ is a bounded increasing sequence, so it has a supremum $A$. Similarly, $\{b_n\}$ is a bounded decreasing sequence, so it has an infimum $B$. Moreover, since $b_n - a_n \to 0$, we have $A=B$. Let $L = A = B$ denote this common value.

Now observe that $L \in [a_n, b_n]$ for every $n$, so in particular $L \in [a,b]$, thus it must be contained in at least one of the open intervals $I$. Since $I$ is open, $I$ contains some neighborhood $N$ of positive length, say $\epsilon >0$, centered at $L$. Now check that $N$ must contain all of the $[a_n, b_n]$ for sufficiently large $n$. So for sufficiently large $n$, each $[a_n, b_n]$ can be covered by a single open interval from the collection, whereas by their construction they can't be covered by finitely many such intervals. Thus we have reached a contradiction.