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**Last_Singularity** Consider the set $\displaystyle B \subseteq [0,1]$ where the elements of $\displaystyle B$ consist of $\displaystyle \frac{p}{\pi q}$ where $\displaystyle \frac{p}{q}$ is rational and between 0 and 1.

Since we define $\displaystyle 0 \leq \frac{p}{q} \leq 1$, then $\displaystyle 0 \leq \frac{p}{\pi q} \leq 1$ as well. That verifies the first requirement.

It is countable because there exists a bijection from the rationals of $\displaystyle [0,1]$ (which are countable) to the set $\displaystyle B$. Specifically, the bijection just takes an element of the rationals of $\displaystyle [0,1]$ and divides it by pi.

It is irrational since there is no way to express any $\displaystyle \frac{p}{\pi q},p \in Z, q \in Z$ as a ratio between two integers.

And it is dense because the rationals are dense, so too will $\displaystyle B$

$\displaystyle QED$

Note that this could have worked with any irrational number, such as $\displaystyle \sqrt{2}$ or $\displaystyle e$ or whatever.

Also note that if a set is dense, it cannot be finite. So by definition, it must be countably infinite or uncountable infinite anyway.