Open Real Interval is not Compact/Normed Vector Space

Proof
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:

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

By Limit of Subsequence equals Limit of Real Sequence, every subsequence of $\sequence {x_n}_{n \mathop \in \N}$ converges to $a$ as well.

However, $a \notin I$.

By definition, $I$ is not compact.