Order relation: inclusion of intervals

Given two any intervals $J$ and $K$ we will say that $J$ is contained in $K$, $J\subseteq K$, if all the elements of $J$ belong to $K$.

The interval $[3,7]$ is included in the interval $\Big(-2,\dfrac{15}{2}\Big)$, and we denote it by: $$[3,7] \subseteq \Big(-2,\dfrac{15}{2}\Big)$$ since $3 \in \Big(-2,\dfrac{15}{2}\Big)$, $7 \in \Big(-2,\dfrac{15}{2}\Big)$ and, consequently, for any $x \in [3,7]$ it is satisfied that $x \in \Big(-2,\dfrac{15}{2}\Big)$

Intuitively, we'd say that this is an order because, given two intervals, it shows which one is bigger: if $J\subseteq K$ then $J$ is smaller than $K$.

In contrast to the order on the real numbers it is not a total order, that is, not every pair of intervals are comparable.

Considering the intervals $(2,3)$ and $(3,4)$, let's see that they are not comparable.

$\dfrac{5}{2}\in (2,3),$ but $\dfrac{5}{2}\notin (3,4),$ therefore it is not true that $(2,3)\subseteq (3,4).$

Likewise,

$\dfrac{10}{3}\in (3,4),$ but $\dfrac{10}{3}\notin (2,3),$ therefore it is not true either that $(3,4)\subseteq (2,3).$

So, they are not comparable.