Closed Real Interval is Compact/Normed Vector Space

Theorem

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

Let $I = \closedint a b$ be a closed real interval.

Then $I$ is compact.

Proof

We have that $\struct {\R, \size {\, \cdot \,}}$ is a normed vector space.

Let $\sequence {a_n}_{n \mathop \in \N}$ be a sequence in $I$.

$\sequence {a_n}$ is bounded by $a$ and $b$.

By Bolzano-Weierstrass theorem, there exists a convergent subsequence $\sequence {a_{n_k}}_{k \in \N}$ with a limit $L$.

Then:

$\forall k \in \N : a \le a_{n_k} \le b$

Take the limit $k \to \infty$:

$a \le L \le b$

Hence, $L \in I$.

The result follows by definition of compact.

$\blacksquare$

Sources

• 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 1.5$: Normed and Banach spaces. Compact sets