# Heine-Borel Theorem/Real Line/Closed and Bounded Interval

## Theorem

Let $\left[{a \,.\,.\, b}\right]$, $a < b$, be a closed and bounded real interval.

Let $S$ be a set of open real sets.

Let $S$ be a cover of $\left[{a \,.\,.\, b}\right]$.

Then there is a finite subset of $S$ that covers $\left[{a \,.\,.\, b}\right]$.

## Proof

Consider the set $T = \left\{ {x \in \left[{a \,.\,.\, b}\right]: \left[{a \,.\,.\, x}\right] \text { is covered by a finite subset of } S}\right\}$.

### Step 1: $a \in T$

It is demonstrated that the number $a$ is an element of $T$.

Consider the interval $\left[ {a \,.\,.\, a} \right]$.

$S$ covers $\left[{a \,.\,.\, a}\right]$ as it covers $\left[{a \,.\,.\, b}\right]$, a superset of $\left[{a \,.\,.\, a}\right]$.

Since $S$ covers $\left[{a \,.\,.\, a}\right]$, there is a set $O$ in $S$ that contains $a$.

We observe:

- $\left\{ {O}\right\}$ is a subset of $S$

- $\left\{ {O}\right\}$ covers $\left[{a \,.\,.\, a}\right]$

By the definition of $T$, $a$ is an element of $T$.

### Step 2: $T$ has Supremum

It is demonstrated that $T$ has a supremum.

We know that $a \in T$.

Therefore, $T$ is non-empty.

Also, $T$ is bounded as $T$ is a subset of $\left[{a \,.\,.\, b}\right]$ which is bounded.

From these two properties of $T$, it follows by the continuum property of the real numbers that $T$ has a supremum, which will be denoted $c$.

### Step 3: $c \in \left[{a \,.\,.\, b}\right]$

It is demonstrated that the number $c$ is an element of $\left[{a \,.\,.\, b}\right]$.

Every element of $T$ is less than or equal to $b$.

So, $b$ is an upper bound for $T$.

Therefore, $c \le b$ as $c$ is the least upper bound of T.

Also, $c \ge a$ because $a \in T$ and $c$ is an upper bound for $T$.

Accordingly, $c \in \left[{a \,.\,.\, b}\right]$.

### Step 4: Finite Cover of $\left[{a \,.\,.\, y}\right]$

It is demonstrated that there exists a finite subset of $S$ covering $\left[{a \,.\,.\, y}\right]$ where $y > c$.

We know that $S$ covers $\left[ {a \,.\,.\, b} \right]$.

Also, $c \in \left[{a \,.\,.\, b}\right]$.

From these two facts follows that there is a set $O_c$ in $S$ that contains $c$.

An $\epsilon \in \R_{>0}$ exists such that $\left({c - \epsilon \,.\,.\, c + \epsilon}\right) \subset O_c$ as $O_c$ is open.

A number $x \in T$ exists in $\left({c - \epsilon \,.\,.\, c} \right)$ as $c$ is the least upper bound of $T$.

By the definition of $T$, a finite subset $S_x$ of $S$ exists such that $S_x$ covers $\left[{a \,.\,.\, x}\right]$ as $x \in T$.

Let $y$ be a number in $\left({c \,.\,.\, c + \epsilon}\right)$.

We observe that $y > c$.

We have that $\left\{{O_c}\right\}$ covers $\left({c - \epsilon \,.\,.\, c + \epsilon}\right)$ as $\left({c - \epsilon \,.\,.\, c + \epsilon}\right) \subset O_c$.

Therefore, $\left\{{O_c}\right\}$ covers $\left[{x \,.\,.\, y}\right]$ as $\left[{x \,.\,.\, y}\right] \subset \left({c - \epsilon \,.\,.\, c + \epsilon}\right)$.

We observe the collection $S_x \cup \left\{{O_c}\right\}$:

- $S_x \cup \left\{{O_c}\right\}$ is a subset of $S$

- $S_x \cup \left\{{O_c}\right\}$ is finite

- $S_x \cup \left\{{O_c}\right\}$ covers $\left[{a \,.\,.\, y}\right]$, the union of $\left[{a \,.\,.\, x}\right]$ and $\left[{x \,.\,.\, y}\right]$

In other words, $S_x \cup \left\{ {O_c}\right\}$ is a finite subset of $S$ covering $\left[{a \,.\,.\, y}\right]$.

### Step 5: $y > b$

It is demonstrated that $y > b$.

We know that $y > c$.

Therefore, $y \notin T$ as $c$ is an upper bound for $T$.

Let us find out exactly why $y \notin T$.

There are two requirements for $y$ to be in $T$:

- $\left[{a \,.\,.\, y}\right]$ is covered by a finite subset of $S$

- $y \in \left[{a \,.\,.\, b}\right]$

The first requirement is satisfied as $S_x \cup \left\{{O_c}\right\}$ is a finite subset of $S$ covering $\left[{a \,.\,.\, y}\right]$.

Therefore, the second requirement is not satisfied as this is the last requirement and the only possibility to satisfy $y \notin T$.

Accordingly, $y \notin \left[{a \,.\,.\, b}\right]$.

Furthermore:

\(\ds y\) | \(>\) | \(\ds c\) | ||||||||||||

\(\ds \iff \ \ \) | \(\ds y\) | \(>\) | \(\ds c \land c \ge a\) | as $c \ge a$ since $c \in \left[{a \,.\,.\, b}\right]$ | ||||||||||

\(\ds \implies \ \ \) | \(\ds y\) | \(>\) | \(\ds a\) | |||||||||||

\(\ds \iff \ \ \) | \(\ds y\) | \(>\) | \(\ds a \land y∉[a..b]\) | as $y \notin \left[{a \,.\,.\, b}\right]$ | ||||||||||

\(\ds \iff \ \ \) | \(\ds y\) | \(>\) | \(\ds b\) |

### Step 6: Finite Cover of $\left[{a \,.\,.\, b}\right]$

It is demonstrated that there exists a finite subset of $S$ covering $\left[{a \,.\,.\, b}\right]$.

We know that $S_x \cup \left\{{O_c}\right\}$ is a finite subset of $S$ covering $\left[{a \,.\,.\, y}\right]$.

Also, $\left[{a \,.\,.\, b}\right]$ is a subset of $\left[{a \,.\,.\, y}\right]$ as $y > b$.

Therefore, $S_x \cup \left\{{O_c}\right\}$ covers $\left[{a \,.\,.\, b}\right]$.

So, $\left[{a \,.\,.\, b}\right]$ is covered by a finite subset of $S$.

$\blacksquare$

## Source of Name

This entry was named for Heinrich Eduard Heine and Émile Borel.

## Sources

- 1988: H.L. Royden:
*Real Analysis*(3rd ed.): Ch. $2$: Section $5$, theorem $15$