# Equivalence of Definitions of Real Interval

## Contents

## Theorem

The following definitions of the concept of **Real Interval** are equivalent:

### Definition 1

A **(real) interval** is a subset $I$ of the real numbers such that:

- $\forall x, y \in I: \forall z \in \R : \paren {x \le z \le y \implies z \in I}$

### Definition 2

A **real interval** is a subset of $\R$ that is one of the following real interval types:

- closed bounded interval
- open bounded interval
- left half-open bounded interval
- right half-open bounded interval
- closed and bounded on the right, also known as a closed unbounded below real interval
- open and bounded on the right, also known as an open unbounded below real interval
- closed and bounded on the left, also known as a closed unbounded above real interval
- open and bounded on the left, also known as an open unbounded above real interval
- unbounded interval without endpoints

## Outline of proof

That an interval according the second definition is an interval according to the first, follows by the properties of the ordering on the real numbers.

To prove the other direction, we distinguish according to if the subset is bounded on the left or on the right, and use the Supremum Principle and Infimum Principle.

## Proof

### 1 implies 2

Let $I\subset \R$ be a subset such that:

- $\forall x, y \in I : \forall z \in \R : \paren {x \le z \le y \implies z \in I}$

Let:

- $a = \map \inf I$ (or $a = -\infty$ if $I$ is not bounded below)
- $b = \map \sup I$ (or $b = +\infty$ if $I$ is not bounded above).

We will prove that:

- $\openint a b \subseteq I \subseteq \closedint a b$

In order not to need to provide a long list of cases, the **temporary** convention is used:

- A square bracket "$[$" or "$]$" next to $\pm \infty$ means the same as a round bracket "$($" or "$)$"
- $\openint a b$ means $\O$ if and only if $a = b$.

Let $z \in \openint a b$.

Then $z \ge a$, so $\exists x \in I: x \le z$ by definition of $a$.

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

Hence by hypothesis $z \in I$.

This proves that $\openint a b \subseteq I$.

From the definitions of $a$ and $b$ it follows that $I \subseteq \closedint a b$.

But if $\openint a b \subseteq I \subseteq \closedint a b$, then $I$ must be one of the various types of real interval.

$\Box$

### 2 implies 1

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

$\blacksquare$

## Sources

- 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $6.2$: Connectedness: Lemma $6.2.6$ - 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 2$: Continuum Property: $\S 2.9$: Intervals