Interval Defined by Betweenness

Lemma

Let $S \subseteq \R$ be a non-empty subset of $\R$.

Then $S$ is an interval of $\R$ iff it satisfies this property:

If $x, y \in S$ and $z \in \R$ such that $x < z < y$, then $z \in S$.

Proof

• If $S$ is a real interval of any of the various kinds, then it clearly has this property by definition.

• Suppose $S$ has this property.

Let:

• $a = \inf \left({S}\right)$ (or $a = -\infty$ if $S$ is not bounded below);
• $b = \sup \left({S}\right)$ (or $b = +\infty$ if $S$ is not bounded above).

We will prove that $\left({a \, . \, . \, b}\right) \subseteq S \subseteq \left[{a \, . \, . \, b}\right]$.

In order not to need to provide lots of lists of cases, we make the temporary conventions:

• A square bracket "$[$" or "$]$" next to $\pm \infty$ means the same as a round bracket "$($" or "$)$";
• $\left({a \, . \, . \, b}\right)$ means $\varnothing$ iff $a = b$.

Let $z \in \left({a \, . \, . \, b}\right)$.

Then $z > a$, so $\exists x \in S: x < z$ by definition of $a$.

Similarly $z < b$, so $\exists y \in S: z < y$ by definition of $b$.

Hence by hypothesis $z \in S$.

This proves that $\left({a \, . \, . \, b}\right) \subseteq S$.

From the definitions of $a$ and $b$ it follows that $S \subseteq \left[{a \, . \, . \, b}\right]$.

But if $\left({a \, . \, . \, b}\right) \subseteq S \subseteq \left[{a \, . \, . \, b}\right]$, then $S$ must be one of the various types of real interval.

$\blacksquare$