Open Real Interval is not Compact/Proof 2

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.

It suffices to demonstrate this for a particular open interval: we use $\openint 0 1$.

Consider the set of open intervals $\openint {\dfrac 1 n} 1$ for all $n \in \Z_{>1}$.

Each of these is an open set in $\openint 0 1$.

Also:

$\openint 0 1 = \ds \bigcup_{n \mathop \ge 2} \openint {\dfrac 1 n} 1$

Thus $\ds \bigcup_{n \mathop \ge 2} \openint {\dfrac 1 n} 1$ is an open cover for $\openint 0 1$.

Note that:

$\forall x \in \openint 0 1: \exists n \in \Z: n > \dfrac 1 x$

and so:

$x \in \openint {\dfrac 1 n} 1$

Consider a finite subcover of $\ds \bigcup_{n \mathop \ge 2} \openint {\dfrac 1 n} 1$.

It will be in the form:

$\set {\openint {\dfrac 1 {n_1} } 1, \openint {\dfrac 1 {n_2} } 1, \ldots, \openint {\dfrac 1 {n_r} } 1}$

But this is just $\openint {\dfrac 1 N} 1$, where $N = \max \set {n_1, n_2, \ldots, n_r}$.

So no finite subcover of $\ds \bigcup_{n \mathop \ge 2} \openint {\dfrac 1 n} 1$ can be a cover for $\openint 0 1$.

$\blacksquare$