# Thread: example of a set

1. ## example of a set

Can anyone think of an example of a set $B$ of real numbers such that $B \subseteq [0,1]$, $B$ is dense, countable and consists entirely of irrational numbers? I can't think of any.

(Let $X$ be a set and let $\leq$ be a total order on $X$. We say that $X$ is dense if for any two $x,y \in X$ with $x < y$ there exists $z \in X$ such that $x)

2. Originally Posted by Jason Bourne
Can anyone think of an example of a set $B$ of real numbers such that $B \subseteq [0,1]$, $B$ is dense, countable and consists entirely of irrational numbers
It is clearly true that such a set must exist.
But I do not know how to give a ‘constructive’ proof of the set.
I hope someone else can do it. I can show it using the axiom of choice.
Here is an outline.
The set of ordered pairs of rational numbers $\left( {p,q} \right)\,,\,0 < p < q < 1\,$ is countable.
Between any two numbers there is an irrational number.
Thus using that we can produce the required set.
Such a set is clearly dense in $\left[ {0,1} \right]$ and must be countable.

3. Originally Posted by Jason Bourne
Can anyone think of an example of a set $B$ of real numbers such that $B \subseteq [0,1]$, $B$ is dense, countable and consists entirely of irrational numbers? I can't think of any.

(Let $X$ be a set and let $\leq$ be a total order on $X$. We say that $X$ is dense if for any two $x,y \in X$ with $x < y$ there exists $z \in X$ such that $x)
Consider the set $B \subseteq [0,1]$ where the elements of $B$ consist of $\frac{p}{\pi q}$ where $\frac{p}{q}$ is rational and between 0 and 1.

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

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

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

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

$QED$

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

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

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

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

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

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

$QED$

Note that this could have worked with any irrational number, such as $\sqrt{2}$ or $e$ 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.
you need to be careful here: if $r_n=\frac{p_n}{q_n} \in [0,1],$ then $\frac{r_n}{\pi} \in [0, 1/\pi]$ and therefore they can't be dense in $[0,1].$ to fix this problem you need to choose $r_n \in [0, \pi].$

5. Thank you for the good catch, NonCommAlg

Sorry for the mistake, Jason Bourne.

6. Originally Posted by Last_Singularity
Thank you for the good catch, NonCommAlg

Sorry for the mistake, Jason Bourne.
No worries, it's easily done. Thanks for all the help.