Real Interval is Bounded in Real Numbers

Theorem
Let $\R$ be the real number line considered as an Euclidean space.

Let $a, b \in \R$.

Let $\mathcal I$ be one of the following real intervals:

where $b \ge a$.

Then $\mathcal I$ is bounded in $\R$.

Proof
Consider the open $\epsilon$-ball $B_\epsilon \left({a}\right)$ where $\epsilon = b + 1 - a$.

As $b \ge a$ we have that $b + 1 > a$ and so $\epsilon > 0$.

Let $x \in \mathcal I$.

Then, whatever type of real interval $\mathcal I$ actually is, $z \ge a$ and $x \le b$.

As $\epsilon > 0$ it follows that $x > a - \epsilon$.

Also:

That is:
 * $a - \epsilon < x < a + \epsilon$

and so:
 * $x \in B_\epsilon \left({a}\right)$

The result follows by definition of bounded space.