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

1. 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 $S$


 * $\left\{ {O} \right\}$ is finite as it contains one element


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

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

2. $T$ has a supremum.

We know that $a \in T$.

Therefore, T is nonempty.

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, c.

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

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

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

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

Consider the collection $S_x \cup \left\{ {O_c} \right\}$.

This collection is finite.

It is a subset of $S$.

Also, it covers $\left[ {a \,.\,.\, y} \right]$ where $y$ is in $\left( {c \,.\,.\, c+\epsilon} \right)$.

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

5. The number $y$ is not an element of $\left[ {a \,.\,.\, b} \right]$.

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

By the definition of $T$, $y \in T$ provided that $y$ also satisfies: $y \in \left[ {a \,.\,.\, b} \right]$.

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

Then $y \in T$ according to the argument above.

This implies that $y \le c$ as $c$ is an upper bound for T.

However, $y \le c$ cannot be true because $y>c$ as $y \in \left( {c \,.\,.\, c+\epsilon} \right)$.

Therefore, $y$ is not an element of $\left[ {a \,.\,.\, b} \right]$.

6. The only possibility for $c$ is $c = b$.

The intersection $\left( {c \,.\,.\, c+\epsilon} \right) \cap \left[ {a \,.\,.\, b} \right]$ is empty as $y$ is in $\left( {c \,.\,.\, c+\epsilon} \right)$ but not in $\left[ {a \,.\,.\, b} \right]$.

The intervals $\left( {c \,.\,.\, c+\epsilon} \right)$ and $\left[ {a \,.\,.\, b} \right]$ overlap when $c \in \left[ {a \,.\,.\, b} \right)$ as they share numbers that are both greater than $c$ and less than $b$.

They do not overlap when $c = b$ as every number in $\left( {c \,.\,.\, c+\epsilon} \right)$ is greater than $b$, the rightmost endpoint of $\left[ {a \,.\,.\, b} \right]$.

Therefore, $\left( {c \,.\,.\, c+\epsilon} \right)$ and $\left[ {a \,.\,.\, b} \right]$ overlap when $c \in \left[ {a \,.\,.\, b} \right)$, but not when $c \in \left\{ {b} \right\}$.

Accordingly, $\left( {c \,.\,.\, c+\epsilon} \right) \cap \left[ {a \,.\,.\, b} \right] = \varnothing$ implies that $c = b$ as $c \in \left[ {a \,.\,.\, b} \right]$, the union of $\left[ {a \,.\,.\, b} \right)$ and $\left\{ {b} \right\}$.

The collection $S_x \cup \left\{ {O_c} \right\}$ covers $\left[ {a \,.\,.\, c} \right]$ as it covers $\left[ {a \,.\,.\, y} \right]$, a superset of $\left[ {a \,.\,.\, c} \right]$.

The fact that $c = b$ means that $S_x \cup \left\{ {O_c} \right\}$ covers $\left[ {a \,.\,.\, b} \right]$.

This finishes the proof as $S_x \cup \left\{ {O_c} \right\}$ is a finite subset of $S$.