Closed Bounded Subset of Real Numbers is Compact

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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 1

A closed and bounded subspace $S$ of $\R$ is a closed subspace of some closed real interval $\left[{a \,.\,.\, b}\right]$.

From Closed Subspace of Compact Space is Compact, it suffices to prove that $\left[{a \,.\,.\, b}\right]$ is compact.


Let $\mathcal U$ be any open cover of $\left[{a \,.\,.\,b}\right]$.

Let:

$G = \left\{{x \in \R: x \ge a, \left[{a \,.\,.\, x}\right] \text{ is covered by a finite subset of } \mathcal U}\right\}$

Let points in $G$ be classified as $\text{good}$ (for $\mathcal U$).

Thus if $z$ is $\text{good}$, then $\left[{a \,.\,.\, z}\right]$ has that finite subcover that is to be demonstrated for the whole of $\left[{a \,.\,.\, b}\right]$.

The aim therefore is to show that $b$ is $\text{good}$.


Let $x$ be $\text{good}$, and $a \le y \le x$.

We have that $\left[{a \,.\,.\, y}\right] \subseteq \left[{a \,.\,.\, x}\right]$.

Thus $\left[{a \,.\,.\, y}\right]$ can be covered with any finite subset of $\mathcal U$ that covers $\left[{a \,.\,.\, x}\right]$.

As $x$ be $\text{good}$, at least one such subset is known to exist.

Thus, by definition, $y$ is $\text{good}$.


Now we show that $G \ne \varnothing$.

At the same time we show that $G \supseteq \left[{a \,.\,.\, a + \delta}\right]$ for some $\delta > 0$.


As $\mathcal U$ covers $\left[{a \,.\,.\, x}\right]$, it follows that $a$ must belong to some $U \in \mathcal U$.

So, let $U \in \mathcal U$ be some open set such that $a \in U$.

Since $U$ is open, $\left[{a \,.\,.\, a + \delta}\right) \subseteq U$ for some $\delta > 0$.

Hence $\left[{a \,.\,.\, x}\right] \subseteq U$ for all $x \in \left[{a \,.\,.\, a + \delta}\right)$.

It follows that all these $x$ are $\text{good}$.

So $G \ne \varnothing$, and:

$(1): \quad G \supseteq \left[{a \,.\,.\, a + \delta}\right]$ for some $\delta > 0$


Now the non-empty set $G$ is either bounded above or it is not.


Suppose $G$ is not bounded above.

Then there is some $c$ which is $\text{good}$ such that $c > b$.

From our initial observation that if $x$ is $\text{good}$, and $a \le y \le x$, then $y$ is $\text{good}$, it follows that $b$ is $\text{good}$, and hence the result.


Now suppose $G$ is bounded above.

By the Continuum Property, $G$ admits a supremum in $\R$.

So let $g = \sup G$.

Let $g > b$.

As $g$ is the least upper bound for $G$, there exists some $c$ which is $\text{good}$ such that $c > b$.

Again, from our initial observation, it follows that $b$ is $\text{good}$, and hence the result.


Aiming for a contradiction, assume that $g \le b$.

From $(1)$ above:

$\left[{a \,.\,.\, a + \delta}\right) \subseteq G$ for some $\delta > 0$

and so $g > a$.

Since $g \in \left[{a \,.\,.\, b}\right]$, $g$ must belong to some $U_0 \in \mathcal U$.

Since $U_0$ is open, there exists some open $\epsilon$-ball $B_\epsilon \left({g}\right)$ of $g$ such that $U_0 \supseteq B_\epsilon \left({g}\right)$.

Since $g > a$, we can arrange that $\epsilon < g - a$.

As $g$ is the least upper bound, there must be a $\text{good}$ $c$ such that $c > g - \epsilon$.

This means $\left[{a \,.\,.\, c}\right]$ is covered by a finite subset of $\mathcal U$, say $\left\{{U_1, U_2, \ldots, U_r}\right\}$.

Then $\left[{a \,.\,.\, g + \dfrac \epsilon 2}\right]$ is covered by $\left\{{U_1, U_2, \ldots, U_r, U_0}\right\}$.

So $g + \dfrac \epsilon 2$ is $\text{good}$, contradicting the fact that $g$ is an upper bound for $G$.

This contradiction implies that $g > b$, and the proof is complete.

$\blacksquare$


Proof 2

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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