Unbounded Set of Real Numbers is not Compact/Normed Vector Space
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
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.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 1.5$: Normed and Banach spaces. Compact sets