Lemma 27.5 in Munkres (Lebesgue number)

Question

I don't understand a few things in the proof below.

First, why does $f(x)\ge \epsilon/n$? Right before this inequality it's stated that $d(x,C_i)\ge \epsilon.$ So each term in the definition of $f(x)$ (namely $\frac{d(x,C_i)}{n}$) is $\ge \epsilon/n$, so shouldn't their sum be $\ge \epsilon$?

Second, why does $\delta \le f(x_0)$ hold?

Finally, I cannot make sense of the very last sentence of the proof.

Answer 1

1). Each $d(x,C_j) \ge 0$, so since $d(x,C_i) \ge \epsilon$, $f(x) \ge \frac{\epsilon}{n}$.

2). $\delta$ is the minimum of $f$, so $f(x_0) \ge \delta$.

3). Since $d(x_0,C_m) \ge \delta$, the delta neighborhood of $x_0$ is disjoint from $C_m$ and thus lies in $A_m$.

Answer 2

Q1: $$f(x)=\frac{1}{n} \sum_{j=1}^n d(x, C_j) \geq \frac{1}{n} d(x, C_i) =\frac{\epsilon}{n}$$

Here we used that a sum of non-negative terms is larger than any of the terms.

Q2 Since $\delta$ is the minimum of $f$, it is smaller than the value of $f$ at any point.