## Saturday, February 5, 2011

### You think there are many rational numbers? You're wrong.

I want to share something very interesting with you that told me my calculus teacher the other day.

Take the rational numbers in the closed interval from 0 to 1. It's easy to see, that this set is given by

$\mathbb{Q} \cap \left [ 0,1 \right ] = \left \{ 0, 1/1, 1/2, 2/2, 1/3, 2/3, 3/3, 1/4, ... \right \}$

So the numbers in this set are countable. Let's call them

$x_0, x_1, x_2, ...$

Now let $\varepsilon > 0$. Cover every $x_n \in \mathbb{Q} \cup \left [ 0,1 \right ]$ with an interval of length $\frac{\varepsilon}{2^n}$, that is

$I_n = \left [ x_n - \frac{\varepsilon}{2^{n+1}}, x_n + \frac{\varepsilon}{2^{n+1}} \right ]$

It follows that the measure of the union of the Intervals fulfills:

$\mu\left ( \bigcup_{n=0}^{\infty} I_n \right ) \leq \sum_{n=0}^{\infty}\frac{\varepsilon}{2^n} = \frac{\varepsilon}{1-\frac{1}{2}} = 2\varepsilon$

So what we see is: if we chose the epsilon small enough, the rational numbers, including the intervals around them, cover nearly no space in the interval from 0 to 1! That means, the space is coverd almost entirely by the irrational numbers.
I think this fact shows us very well what the difference between "countable" and "uncountable" sets is, and how much more irrational numbers than rational numbers there are even though they are both infinite many.