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### Nonoverlapping Intervals

```
Date: 04/08/99 at 22:47:22
From: Kai - Yuan Cheng
Subject: Uncountable number

Show that there can not be an uncountable (nondenumerable) number of
nonoverlapping intervals on the real line.

Thanks.
```

```
Date: 04/09/99 at 11:23:22
From: Doctor Floor
Subject: Re: Uncountable number

Hi, Kai-Yuan Cheng,

First we have to be sure that the number of rational numbers is
countable. You might like to read why that is true in the Dr. Math
archives:

http://mathforum.org/dr.math/problems/stephanie9.9.98.html

Then we note that any (non-empty) interval on the real line contains a
rational number. Or, in other words, that between any two unequal real
numbers there is a rational number. To see that, think of the two real
numbers in decimal progression, e.g.:

0.12491287491827409182734091827342341235123519...
0.12491287491827409182734091827342341235123520...

There must be a point where the two progressions become different,
which is the last decimal shown in the example.

It is clear that we can construct a rational number between these two.
There seems only to be a problem when the first progression would go
on with 9's only, and the second with 0's only. But then the two
numbers are equal.

So we have that in any interval there is a rational number. Since the
intervals are nonoverlapping, it follows that there cannot be more of
those intervals than there are rational numbers. As a consequence of
the countability of the rational numbers, the number of nonoverlapping
intervals on the real line cannot be uncountable.

Best regards,

- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Analysis

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