How to show that $[c_i, d_i] \subset [a_i- \lambda, b_i + \lambda]$?

Question

In the book of Analysis on Manifolds by Munkres, at page 152, it is given that

However, I cannot see how do we we have that $[c_i, d_i] \subset [a_i- \lambda, b_i + \lambda]$. I mean I have tried forming counterexamples, but every case what obeying the above condition, but still I cannot see the logical implication.

So, why do we have $[c_i, d_i] \subset [a_i- \lambda, b_i + \lambda]$, and how can we show it ?

Edit:

Answer 1

Note that "$1/N$ is less than the small of $\delta$ and $\lambda$.", In the figure below, note that the interval $[a_i - 1/N,a_i + 1/N]$ contains both neighboring elements of the form $m/N$. That is, $$ a_i - \lambda < a_i - 1/N \leq c_i $$ and similarly, $$ b_i + \lambda > b_i + 1/N \geq d_i $$

Edit:

Note that the maximality of $c_i$ contradicts with that if $c_i + 1/N \leq a_i$, so we have $c_i \leq a_i \leq c_i + 1/N$.

Answer 2

By definition, $c_i$ is the largest number of the form $\frac{m}{N}$ such that $c_i \leqslant a_i$. So if we take any larger number of that form, e.g. $\frac{m+1}{N}$, it must be larger than $a_i$, i.e. $a_i < \frac{m+1}{N} = c_i + \frac{1}{N}$. Now since $\frac{1}{N} \leqslant \lambda$, we have that

$$a_i - \lambda \leqslant a_i - \frac{1}{N} < c_i.$$

In an analogous way we prove that $d_i < b_i + \lambda$, which gives $[c_i, d_i] \subsetneq [a_i - \lambda, b_i + \lambda]$.