Open Real Interval is not Compact/Normed Vector Space

Theorem

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

Let $I = \openint a b$ be an open real interval.

Then $I$ is not compact.

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

Also, $I$ is bounded by $a$ and $b$.

Consider a sequence $\ds \sequence {x_n}_{n \mathop \in \N}$ with $\ds x_n = a + \frac {b - a} {2n}$.

Let $\epsilon \in \R_{\mathop > 0}$.

Let $\ds N = \frac {b - a} {2 \epsilon}$.

Then for all $n \in \N$ such that $n > N$ we have that:

$\ds \size {a + \frac {b - a} {2n} - a}$

$=$

$\ds \size {\frac {b - a} {2n} }$

$\ds $

$<$

$\ds \frac {b - a} {2N}$

$b > a$

$\ds $

$=$

$\ds \frac {b - a} 2 \frac {2 \epsilon} {b - a}$

$\ds $

$=$

$\ds \epsilon$

Hence, $\sequence {x_n}_{n \mathop \in \N}$ converges to $a$.

However, $a \notin I$.

By definition, $I$ is not compact.

$\blacksquare$

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