Difference of Unbounded Closed Intervals

Theorem

Let $a, b \in \R$ have $a < b$.

Then:

$\hointl {-\infty} b \setminus \hointl {-\infty} a = \hointl a b$

where $\setminus$ denotes set difference.

Proof

Note that:

$x \in \hointl {-\infty} b \setminus \hointl {-\infty} a$

if and only if:

$x \in \hointl {-\infty} b$ but $x \not \in \hointl {-\infty} a$.

That is:

$x \le b$ but it is not the case that $x \le a$.

So this is equivalent to:

$x \le b$ and $x > a$.

That is:

$x \in \hointl a b$

So:

$\hointl {-\infty} b \setminus \hointl {-\infty} a = \hointl a b$

$\blacksquare$