I have a so easy question. I have done it's answer by myself. I want you to only check my answer please. Does there exist any mistake or the missing?
The set of irrational numbers – $\Bbb Q$ is neither open nor closed.
Proof:
Assume $\Bbb Q$ were open. There would be a neighborhood of $0$, and so an interval containing $0$ lying entirely within $\Bbb Q$. However, each such interval contains irrational numbers, which is a contradiction.
suppose $\Bbb Q$ were closed. $\Bbb R-\Bbb Q$ is open. There is a neighborhood of $\pi$ and therefore an interval containing $\pi$ lying completely within $\Bbb R-\Bbb Q$ . however again each such interval contains rational numbers, which is a contradiction.
Best Answer
That is fine, yes, though using $\sqrt 2$ instead of $\pi$ would make the proof not rely on knowing the latter was irrational!
Also perhaps you could check that you know why there are rationals/irrationals in every open interval.