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The theorem, as seen in Analysis 1 textbook by Vladimir A. Zorich:

Every family of open intervals, that covers a closed interval, contains a finite subfamily, that covers the closed interval.

Proof. Let $S=\{U\}$ be a family of open intervals $U$ which cover the closed interval $[a,b]=I_1$. If $I_1$ cannot be covered by a finite set of intervals of the family $S$, then we divide $I_1$ in two halves. At least one of the halves, we denote with $I_2$, allows no finite coverage. We repeat this process with the interval $I_2$ and so on.

In doing so, we create a nested sequence $I_1\supset I_2 \supset \dots \supset I_n \supset \dots$ of closed intervals, amongst which none allow a coverage of a finite subfamily of S. Since the length of $I_n$ is equal to $|I_n|=|I_1|\cdot 2^{-n}$, the sequence $\{I_n\}$ contains intervals of arbitrarily small length. According to the Nested Interval Property, there exists a point $c$, which is in all of these intervals $I_n, n\in \mathbb{N}$. Since $c \in I_1 = [a,b]$, there exists an open interval $ (\alpha, \beta)=U \in S$, that contains $c$, i.e., $\alpha < c < \beta$. Let $\epsilon=min\{c-\alpha, \beta - c\}$. In the previously created sequence of intervals, we can find an interval $I_n$, such that $|I_n|< \epsilon$. Since $c \in I_n$ and $|I_n|<\epsilon$, it follows that $I_n \subset U=(\alpha, \beta)$. This is in contradiction with the fact that the interval $I_n$ cannot be covered with a finite set of intervals of the family. And therefore the initial statement is true.

End of proof.


Two things I fail to grasp:

  1. Why is the choice of $\epsilon=min\{c-\alpha, \beta - c\}$ a good one, and how are you supposed to come up with it yourself? Or what piece of information we had before the choice of $\epsilon$, is supposed to indicate what the choice should be?
  2. Why is from $c\in I_n$ and $|I_n|<\epsilon $ following that $I_n\subset U=(\alpha, \beta)$ ?

I translated the text from german, I hope there are no discrepancies between the terms.

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$c-\alpha$ is the distance between $c$ and the lower end of the interval $(\alpha,\beta)$. Similarly, $\beta-c$ is the distance from $c$ to the upper end of the interval. Normally we would write $\vert \alpha-c\vert$ and $\vert \beta-c\vert$, but since we know that $\alpha<c<\beta$, we can leave out the absolute values and just choose the correct order for the subtraction: $c-\alpha$ because $\alpha<c$, and $\beta-c$ because $c<\beta$. And then the minimum $\epsilon$ of the two is just the minimum distance of $c$ to the interval boundaries. Meaning that anything that's closer than $\epsilon$ to $c$ is both larger than $\alpha$ and smaller than $\beta$, so everything that's within the distance $\epsilon$ of $c$ is also inside the interval $(\alpha,\beta)$. And that's the case for $I_n$: since it contains $c$ and has length smaller than $\epsilon$, all points in $I_n$ are closer than $\epsilon$ to $c$, and are thus contained in $(\alpha,\beta)$. And then so is $I_n$.

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  • $\begingroup$ Perfectly clear now. Many thanks $\endgroup$
    – tau20
    Aug 20 '20 at 17:36

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