Unbounded Set of Real Numbers is not Compact
Theorem
Let $\R$ be the set of real numbers considered as a Euclidean space.
Let $S \subseteq \R$ be unbounded in $\R$.
Then $S$ is not a compact subspace of $\R$.
Proof 1
By the rule of transposition, it suffices to show that if $S$ is a compact subspace of $\R$, then $S$ is bounded.
Let $\CC$ be the set of all open $\epsilon$-balls of $0$ in $\R$:
- $\CC = \set {\map {B_\epsilon} 0: \epsilon \in \R_{>0}}$
We have that:
- $\ds \bigcup \CC = \R \supseteq S$
From Open Ball of Metric Space is Open Set, it follows that $\CC$ is an open cover for $S$.
Let $\FF$ be a finite subcover of $\CC$ for $S$.
Then $\ds \bigcup \FF$ is the largest open $\epsilon$-ball in $\FF$.
Thus:
- $S \subseteq \ds \bigcup \FF \in \CC$
Hence, $S$ is bounded.
$\blacksquare$
Proof 2
From:
- Real Number Line is Metric Space
- Compact Metric Space is Totally Bounded
- Totally Bounded Metric Space is Bounded
the result follows by the rule of transposition.
$\blacksquare$
Proof 3
We have that $\struct {\R, \size {\, \cdot \,} }$ is a normed vector space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence such that $x_n = n$.
Aiming for a contradiction, suppose $\sequence {x_n}_{n \mathop \in \N}$ poseses a convergent subsequence $\sequence {x_{n_k}}_{k \mathop \in \N}$.
By Convergent Sequence is Cauchy Sequence, $\sequence {x_{n_k}}_{k \mathop \in \N}$ is Cauchy.
However:
| \(\ds \forall k, m, n \in \N: \, \) | \(\ds \size {x_{n_k} - x_{n_m} }\) | \(\ge\) | \(\ds \size {x_{n_{m + 1} } - x_{n_m} }\) | |||||||||||
| \(\ds \) | \(\ge\) | \(\ds \size {x_{n + 1} - x_n }\) | ||||||||||||
| \(\ds \) | \(\ge\) | \(\ds \size {n + 1 - n}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1\) |
This contradicts the definition of Cauchy sequence.
Hence, there is no convergent subsequence of $\sequence {x_n}_{n \mathop \in \N}$.
By definition, $\struct {\R, \size {\, \cdot \,} }$ is not compact.
$\blacksquare$