Prove that the irrationals are dense in the irrationals

Question

I'm having trouble with this proof. I think I need to show that every neighborhood of an irrational number contains at lest one other irrational number. I'm getting lost in the notation.

Answer 1

A cardinality argument works. Any $\epsilon$-neighborhood of a point contains uncountably many points, whereas the rationals are countable. If any such neighborhood contains only rational numbers, its cardinality is at most countable, a contradiction.

To see that any neighborhood $(x-\epsilon,x+\epsilon)$ contains uncountably many points, map it linearly to $(-\frac{\pi}{2}, \frac{\pi}{2})$ and apply the arctan function to map that interval to $\mathbb{R}$ itself.

Answer 2

Any set is dense in itself, as $\bar A\supset A$.