Heine-Borel Theorem/Real Line/Closed and Bounded Set

Theorem
Let $F$ be a closed and bounded real set.

Let $C$ be a collection of open real sets.

Let $C$ be a cover of $F$.

Then there is a finite subset of $C$ that covers $F$.

Proof
We know that $C$ is a collection of open real sets that covers $F$.

In other words, $C$ is an open cover of $F$.

We need to show that there is a finite subset of $C$ that covers $F$.

In other words, we need to show that $C$ has a finite subcover.

Let $F_o$ be the complement of $F$.

By the definition of closed real set, $F_o$ is open as $F$ is closed.

1. $C^*$ is an open cover of $\left[ {a \,.\,.\, b} \right]$.

Since $F$ is bounded, $F$ is contained in a closed and bounded interval $\left[ {a \,.\,.\, b} \right]$ where $a,b \in \R$.

Define $C^* = C \cup \left\{ {F_o} \right\}$.

Like $C$, $C^*$ is a collection of open real sets as $F_o$ is open.

$C^*$ covers $F \cup F_o$ as $C$ covers $F$ and $\left\{ {F_o} \right\}$ covers $F_o$.

$F_o \cup F$ equals $\R$ as $F_o$ is the complement of $F$.

So, $C^*$ covers $\R$.

Furthermore, $C^*$ is an open cover of $\left[ {a \,.\,.\, b} \right]$ as $\left[ {a \,.\,.\, b} \right]$ is a subset of $\R$.

2. $C^*$ has a finite subcover $C^*_f$.

$C^*$ is an open cover of the closed and bounded real interval $\left[ {a \,.\,.\, b} \right]$.

Therefore, by Open Cover of Closed and Bounded Real Interval has Finite Subcover, $C^*$ has a finite subcover $C^*_f$.

3. $C$ has a finite subcover $C_f$.

Note that $F_o$ is the only element in $C^*$ that is not an element of $C$.

Therefore, $F_o$ is the only possible element in $C^*_f$ that is not an element of $C$ as $C^*_f$ is a subset of $C^*$.

This means that $C^*_f$ \ $\left\{ {F_o} \right\}$ is a subset of $C$.

Define $C_f$ = $C^*_f$ \ $\left\{ {F_o} \right\}$.

According to the reasoning above, $C_f$ is a subset of $C$.

Also, $C_f$ is finite as $C^*_f$ is finite.

What remains is to show that $C_f$ covers $F$.

We have:

Furthermore, as $F \subseteq \left[ {a \,.\,.\, b} \right]$ is true, we get:

Thus, $C_f$ covers $F$.

This finishes the proof.