How are the following inequalities concluded based on this first one?

Question

$$I-\frac{\epsilon}{3} \leq s(f,T) \leq \underline{I} \leq \overline{I}\leq S(f,T) \leq I+ \frac{\epsilon}{3}$$

from this, the following is concluded, but how? $$1.\ \ \ 0 \leq |I-\underline{I}|\leq \frac{2 \epsilon}{3} $$ $$2.\ \ \ 0 \leq |I-\overline{I}|\leq \frac{2 \epsilon}{3}$$ $$3.\ \ \ 0 \leq |\overline{I}-\underline{I}|\leq \frac{2 \epsilon}{3}$$ $$4.\ \ \ 0 \leq S(f,T)-s(f,T)\leq \frac{2 \epsilon}{3}$$ $$5.\ \ \ 0 \leq |I-s(f,T)|\leq \frac{2 \epsilon}{3}$$ $$6.\ \ \ 0 \leq |I-S(f,T)|\leq \frac{2 \epsilon}{3}$$

The first two I understand.

Answer 1

Note that the two end elements in the inequality chain are $\frac {2\epsilon}3$ apart. That means te distance between any two elements in the chain are at most $\frac {2\epsilon}3$ apart.