Cover of Interval By Closed Intervals is not Pairwise Disjoint

Theorem

Let $[a .. b]$ be a closed interval in $\R$.

Let $\mathcal J$ be a collection of two or more closed intervals contained in $[a .. b]$ such that $\displaystyle \bigcup {\mathcal J} = [a .. b]$.

Then the intervals in $\mathcal J$ are not pairwise disjoint.

Proof

Suppose that the intervals of $\mathcal J$ are pairwise disjoint.

Let $I = [p .. q]$ be the unique interval of $\mathcal J$ such that $a \in I$.

Let $J = [r .. s]$ the unique interval containing the least real number not in $I$.

These choices are possible since $\mathcal J$ has at least two elements, and they are supposed disjoint.

If $q = r$ then $I \cap J \neq \varnothing$, a contradiction.

If $r < q$ then $I \cap J \neq \varnothing$, a contradiction.

If $r > q$ then there is some real number with $q < \alpha < r$.

Therefore $\alpha \notin I$.

Since $J$ contains the least real number not in $I$ it follows that there is $\beta \in J$ with $\beta < \alpha$.

But we also have that $\alpha < r \leq \beta$ for all $\beta$ for all $\beta \in J$, a contradiction.

This exhausts all the possibilities, and we conclude that the intervals of $\mathcal J$ are not pairwise disjoint.

$\blacksquare$