Real Interval is Bounded in Real Numbers
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Theorem
Let $\R$ be the real number line considered as an Euclidean space.
Let $a, b \in \R$.
Let $\II$ be one of the following real intervals:
\(\ds \openint a b\) | \(:=\) | \(\ds \set {x \in \R: a < x < b}\) | Open Real Interval | |||||||||||
\(\ds \hointr a b\) | \(:=\) | \(\ds \set {x \in \R: a \le x < b}\) | Half-Open (to the right) Real Interval | |||||||||||
\(\ds \hointl a b\) | \(:=\) | \(\ds \set {x \in \R: a < x \le b}\) | Half-Open (to the left) Real Interval | |||||||||||
\(\ds \closedint a b\) | \(:=\) | \(\ds \set {x \in \R: a \le x \le b}\) | Closed Real Interval |
where $b \ge a$.
Then $\II$ is bounded in $\R$.
Proof
Consider the open $\epsilon$-ball $\map {B_\epsilon} a$ where $\epsilon = b + 1 - a$.
As $b \ge a$ we have that $b + 1 > a$ and so $\epsilon > 0$.
Let $x \in \II$.
Then, whatever type of real interval $\II$ actually is, $z \ge a$ and $x \le b$.
As $\epsilon > 0$ it follows that $x > a - \epsilon$.
Also:
\(\ds x\) | \(\le\) | \(\ds b\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a + b - a\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds a + \paren {b + 1 - a}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a + \epsilon\) |
That is:
- $a - \epsilon < x < a + \epsilon$
and so:
- $x \in \map {B_\epsilon} a$
The result follows by definition of bounded space.
$\blacksquare$
Sources
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{III}$: Metric Spaces: Compactness