Covering rational numbers by countable open intervals

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September 26th 2009, 05:29 PM
sabbatai

Covering rational numbers by countable open intervals

Hi,
I have a question asking to show that rational numbers in the closed interval [0;1] can be covered by a countable family of open intervals of total length less or equaly to 1/2.

I have some understanding, but I don't know how to solve the issue.

For the question, I thought that as the initial interval is close we can always find a smaller interval (or a disk) that has its center at the boundary point of the interval. Thus, as the interval is closed, we cannot find infinitely smaller disk that are contained in the interval.

Would you guys help me to formalize my thoughts and could you give me some ideas?
September 26th 2009, 05:42 PM
Sampras

nested interval theorem.
September 26th 2009, 05:43 PM
sabbatai

Quote:

Originally Posted by Sampras

nested interval theorem.

I don't think it is nested interval theorem. nested interval theorem involves countable closed intervals, this one asks for countable open intervals
September 26th 2009, 06:44 PM
aliceinwonderland

Quote:

Originally Posted by sabbatai

Hi,
I have a question asking to show that rational numbers in the closed interval [0;1] can be covered by a countable family of open intervals of total length less or equaly to 1/2.

I have some understanding, but I don't know how to solve the issue.

For the question, I thought that as the initial interval is close we can always find a smaller interval (or a disk) that has its center at the boundary point of the interval. Thus, as the interval is closed, we cannot find infinitely smaller disk that are contained in the interval.

Would you guys help me to formalize my thoughts and could you give me some ideas?

Any subset of rational numbers including itself is of measure zero. Indeed any countable set is of measure zero.

Let $A = \{x_1, x_2, x_3, ... \}$ be a set of rational numbers in [0, 1]. For each positive integer n, choose an open interval $I_n$ containing $x_n$ such that $|I_n| < \frac{\epsilon}{3^n}$ . An infinite union of $I_n$ covers A, i.e $A \subset \bigcup_{n=1}^{\infty}I_n$ .
Since $\sum_{n=1}^{\infty}|I_n| \leq \epsilon \sum_{n=1}^{\infty}\frac{1}{3^n} = \frac{\epsilon}{2} < \epsilon$ , A is of measure zero.

To get an intuitive idea, just pick a rational point p. Find an arbitrary small open interval containing p. For example, consider an open interval containing p, (p-1/k, p, p+1/k), where 1/k goes to an arbitrarily small number. The length of an open interval containg p can go to an arbitrarily small, but it still contains a point p. As shown above, a countable summation of the length of these arbitrarily small open intervals still remain arbitrarily small.