Proof that no segment is compact Does anyone have a simple proof (without using any theorems of compactness) that no segment of the form $(a, b)$ in $\mathbb{R}$ is compact?
Definitions:  For any subset $E$ of a metric space $X$, an open cover is a collection of sets $\{G_\alpha\}$ which are open in $X$, such that $E \subset \bigcup_\alpha G_\alpha$.  $E$ is compact if every one of its open covers has a finite open subcover.
Proof so far:  Let $G_n = (a + \frac{1}{n}, b - \frac{1}{n})$.  Then $\bigcup_{i=1}^n G_n = (a, b)$.  [Is there a concise way to prove this fact?]  So $G = \{G_n | n \in \mathbb{N}\}$ is an open cover of $(a, b)$, but $G$ has no finite open subcover.  [Why is that true?]
 A: Since no one gave a full answer, here's a proof using the Archimedean Property of $\mathbb{R}$.  Thanks for the original comments, and please let me know if anything here is wrong!
To show that $(a, b)$ is not compact, we just need one example
of an open cover that has no finite open subcover.
We will use the cover
$$\{ G_n \} = \left\{ \left( a + \frac{1}{n}, b - \frac{1}{n} \right) \mid n \in \mathbb{N} \right\}$$
(If this gives us an invalid segment such as $(1, 0)$, treat it an empty element.)
Why does $\bigcup_{n=1}^\infty G_n$ cover $(a, b)$?
Well, for any element $x \in (a, b)$, the Archimedean Property provides an $n \in \mathbb{N}$
such that $n > \max \{ \frac{1}{x - a}, \frac{1}{b - x} \}$.  Then
$$nx - na > 1 \text{ and } nb - nx > 1 \Rightarrow na + 1 < nx < nb - 1 \Rightarrow x \in \left( a + \frac{1}{n}, b - \frac{1}{n} \right).$$
Thus every element of $(a, b)$ is in $G_n$ for some $n \in \mathbb{N}$,
so $(a, b) \subset \bigcup_{n=1}^\infty G_n$.
Why does $\{ G_n \}$ have no finite subcover?  Well, for any $i > j$,
$a + \frac{1}{i} < a + \frac{1}{j} < b - \frac{1}{j} < b - \frac{1}{i}$, so that $G_i \supset G_j$.
Thus for any $k \in \mathbb{N}$ with $k < \infty$,
$\bigcup_{n=1}^k G_n = G_k = (a + \frac{1}{k}, b - \frac{1}{k})$.
But $(a, b) \not\subset G_k$---since, for example, $a + \frac{1}{2k} \in (a, b)$
but $\notin (a + \frac{1}{k}, b - \frac{1}{k})$.
