# Help with Proof of Borel-Lebesgue theorem

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.

$$c-\alpha$$ 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 $$\alpha, we can leave out the absolute values and just choose the correct order for the subtraction: $$c-\alpha$$ because $$\alpha, 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 $$I_n$$.

• Perfectly clear now. Many thanks Aug 20 '20 at 17:36