Unbounded Set of Real Numbers is not Compact/Normed Vector Space

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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.

$\blacksquare$


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