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First thank you in advance. I am learning real analysis and come across a theorem which states

$\mathbb{Q}$ can be hidden by an arbitrarily small piece of line. Let $\mathbb{Q} =\{r_1,...,r_n,...\}$ where we can cover $r_1 by \frac{\epsilon}{2}$ and cover $r_2 by \frac{\epsilon}{4}$ ... and $r_n by \frac{\epsilon}{2^{n}}$.

what does this mean?

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  • $\begingroup$ What do you mean by "what does it mean"? Are you having trouble understanding what the theorem is talking about? Are you looking for intuitive ramifications? For one, we can use this fact to find closed sets containing no isolated points and no rationals. $\endgroup$ – Omnomnomnom Sep 10 '14 at 4:25
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This means that for any positive real number $\epsilon$, there exists a sequence of open intervals $(a_n, b_n)$ such that $\sum_{n = 1}^\infty b_n - a_n \leq \epsilon$ and $\mathbb{Q} \subseteq \bigcup_{n \in \mathbb{N}} (a_n, b_n)$.

This is true since if one fixed a enumeration $\{q_n : n \in \mathbb{N}\}$ of $\mathbb{Q}$, then let $a_n = q_n - \frac{\epsilon}{2^{n + 1}}$ and $b_n = q_n + \frac{\epsilon}{2^{n + 1}}$. It is clear that $q_n \in (a_n, b_n)$. Hence $\mathbb{Q} \subseteq \bigcup_{n \in \mathbb{N}} (a_n, b_n)$. Moreover, $b_n - a_n = \frac{\epsilon}{2^n}$. Finally, $\sum_{i = 1}^n \frac{\epsilon}{2^n} = \epsilon$ since this is a convergent geometric series.

Intuitively, $\mathbb{Q}$ can be covered by a set whose "length" is less than $\epsilon$. Note that since $\epsilon$ is arbitrary, this intuitively means that the set $\mathbb{Q}$ has "length" $0$.

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  • $\begingroup$ is the sequence of open intervals in the reals? a_n in reals? same for b ? i believe so... $\endgroup$ – sophie-germain Sep 10 '14 at 5:42
  • $\begingroup$ @AnthonyColombo Yes. The term "line" means the real numbers $\mathbb{R}$. We think of $\mathbb{Q}$ inside of $\mathbb{R}$ in the natural way. $\endgroup$ – William Sep 10 '14 at 5:44
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This means that the rational numbers, (in the scope of the reals), have zero length, (or measure)

If we cover, (i.e.), place an interval on top of each rational number $r_n$ of length $$\frac{\epsilon}{2^n}$$
Then the sum of all the lengths of the intervals will be: $$ \sum_{n \in \mathbb{N}} \frac{\epsilon}{2^n} = \epsilon \sum_{n\in \mathbb{N}}\frac{1}{2^n} = \epsilon\cdot 1 = \epsilon $$

This is for arbitrarily small positive numbers $\epsilon$, for example think of $\epsilon = .0000000000000000001$, (or smaller), (ten trillion, billion, quadrillion zero's and then a $1$).

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  • $\begingroup$ downvote???????? $\endgroup$ – Rustyn Sep 11 '14 at 16:45

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