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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?]

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The first assertion follows from the Archimedean property and the second assertion follows from the fact that the union of any finite subcover of $\{ G_n \}$ is contained in $G_n$ for some $n$ (the maximum $n$ which appears in the subcover). – Qiaochu Yuan Feb 14 '12 at 0:51
If you don't see intuitively why the proof works, draw a picture. In topology, drawing a picture almost always helps immensely. – Lopsy Feb 14 '12 at 0:53
I just found an answer to the question "[Is there a concise way to prove this fact?]" For every $x \in (a, b)$, the Archimedean Property of $\mathbb{R}$ asserts that there exists an $n \in \mathbb{N}$ such that $n > \max \{\frac{1}{x-a}, \frac{1}{b-x}\}$, so that $na + a < nx < nb - 1 \Rightarrow a + \frac{1}{n} < x < b - \frac{1}{n} \Rightarrow x \in G_n$. Thus for every $x \in (a, b)$, $x \in \bigcup_{i=1}^n G_n = G$. – jamaicanworm Feb 14 '12 at 0:53
Qiaochu, you don't need the Archimedian property to know that $(a,b)$ is not compact. Rather, you just need to know that the order is dense. Even non-Archimedian dense orders have open intervals non-compact, since $(a,b)=\bigcup_{a\lt c\lt d\lt b}(c,d)$ is an open cover with no finite subcover. – JDH Feb 14 '12 at 1:31
Yes, I agree, you need only density. The $\mathbb{R}$ order is dense because we can take averages. (So we don't need to deal with the Archimedian property.) – JDH Feb 14 '12 at 1:44
up vote 1 down vote accepted

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 $\notin (a + \frac{1}{k}, b - \frac{1}{k})$.

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