I'm working through the definitions given in 2.18 in Baby Rudin, and I was wondering how to prove that $E = [0,1] \cap \Bbb{Q}$ is closed in $\Bbb{Q}$.
Clearly, the set $[0,1] \cap \Bbb{R}$ is closed in $\Bbb{R}$, since for all neighborhoods of a point $p \in [0,1]$, we can always find a $q \neq p$ such that $q \in [0,1]$. Intuitively, this is because for all $p \in [0,1] \cap \Bbb{R}$, we can always find a $q \in [0,1] \cap \Bbb{R}$ such that $p - \epsilon < q < p$ or $p < q < p+\epsilon$ for all $\epsilon >0$.
However, that the set $E = [0,1] \cap \Bbb{Q}$ is closed in $\Bbb{Q}$ is sort of clear, but not intuitive to me, since $\Bbb{Q}$ is not uncountably infinite as $\Bbb{R}$ is. Does one have to use that fact that it is closed in $\Bbb{R}$ and that $\Bbb{Q}$ is dense in $\Bbb{R}$? How would one solve it from first principles? i.e. Something along the lines of: Consider an arbitrary $p \in [0,1] \cap \Bbb{Q}$, we show that it is a limit point by...
Any help would be greatly appreciated!