Closed Bounded Subset of Real Numbers is Compact/Proof 2

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Theorem

Let $\R$ be the real number line considered as an Euclidean space.

Let $S \subseteq \R$ be a closed and bounded subspace of $\R$.


Then $S$ is compact in $\R$.


Proof

Let $S$ be closed and bounded.

As $S$ is bounded, there exist some $a, b \in \R$ such that:

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

where $\left({a \,.\,.\, b}\right)$ is the open interval between $a$ and $b$.

It follows that $S \subseteq \left[{a \,.\,.\, b}\right]$.


Consider the set:

$U = \complement_\R \left({S}\right) \cap \left({a - 1 \,.\,.\, b + 1}\right)$

By inspection it can be seen that:

$U = \left({a - 1 \,.\,.\, a}\right) \cup \left({b \,.\,.\, b + 1}\right)$

By Union of Open Sets of Metric Space is Open, it follows that $U$ is open in $\R$.


Let $\mathcal C$ be an open cover for $S$.

Then we can construct the open cover $\mathcal C' = \mathcal C \cup \left\{{U}\right\}$ for $\left[{a \,.\,.\, b}\right]$.

Let $\mathcal F' \subseteq \mathcal C'$ be a finite subcover of $\mathcal C'$ for $\left[{a \,.\,.\, b}\right]$.

Then $\mathcal F = \mathcal F' \setminus \left\{{U}\right\}$ is the desired finite subcover for $S$.

Hence it is sufficient to prove that any open cover for $\left[{a \,.\,.\, b}\right]$, has a finite subcover.


So, suppose $S = \left[{a \,.\,.\, b}\right]$ and create the set $T \subseteq \left[{a \,.\,.\, b}\right]$ as follows:

Let $a \le x \le b$.

Then $x \in T$ if and only if $\left[{a \,.\,.\, x}\right]$ has a finite subcover of $\mathcal C$.

We have that $a \in T$ and that $b$ is an upper bound for $T$.

Let $l = \sup \left({T}\right)$ be the supremum of $T$.

Let $L \in \mathcal C$ such that $l \in L$.

Since $L$ is open:

$\exists \epsilon \in \R_{>0}: \left({l - \epsilon \,.\,.\, l + \epsilon}\right) \subseteq L$

Since $l = \sup \left({T}\right)$ there exists $t \in T$ such that $t > l - \epsilon$.

We have that $\mathcal F$ is a finite subset of $\mathcal C$ such that $\displaystyle \left[{a \,.\,.\, t}\right] \subseteq \bigcup \mathcal F$.

Then $\mathcal F \cup \left\{{L}\right\}$ is a finite subset of $\mathcal C$ whose union contains $\left[{a \,.\,.\, l + \delta}\right]$ for every $\delta \in \left({0 \,.\,.\, \epsilon}\right)$.

Since $l$ is an upper bound for $T$, it follows that $l + \delta \notin T$.

Thus $l + \delta > b$ for all $\delta > 0$.

That is, $l \ge b$.

But by definition, $l \le b$.

So $l = b$ and so $\mathcal C$ has a finite subcover for $\left[{a \,.\,.\, b}\right]$.

$\blacksquare$


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