Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?]

share|improve this question
3  
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
1  
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
1  
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
3  
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
1  
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

1 Answer 1

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.