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:

 $\displaystyle \openint a b$ $:=$ $\displaystyle \set {x \in \R: a < x < b}$ Open Real Interval $\displaystyle \hointr a b$ $:=$ $\displaystyle \set {x \in \R: a \le x < b}$ Half-Open (to the right) Real Interval $\displaystyle \hointl a b$ $:=$ $\displaystyle \set {x \in \R: a < x \le b}$ Half-Open (to the left) Real Interval $\displaystyle \closedint a b$ $:=$ $\displaystyle \set {x \in \R: a \le x \le b}$ Closed Real Interval

where $b \ge a$.

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

Proof

Consider the open $\epsilon$-ball $B_\epsilon \paren a$ 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:

 $\displaystyle x$ $\le$ $\displaystyle b$ $\displaystyle$ $=$ $\displaystyle a + b - a$ $\displaystyle$ $<$ $\displaystyle a + \left({b + 1 - a}\right)$ $\displaystyle$ $=$ $\displaystyle a + \epsilon$

That is:

$a - \epsilon < x < a + \epsilon$

and so:

$x \in B_\epsilon \paren a$

The result follows by definition of bounded space.

$\blacksquare$