If we consider the rationals, $\mathbb{Q}$, as a metric space with the usual metric of the real line, then is the set $B=\{q\in\mathbb{Q} \mid {q^2}\lt 2\}$ a closed set? I know the definition of a closed set is a set that contains all its limit points. And I understand that here $\sqrt{2}$ is a limit point but it's not in our set but it's also not in the rationals so it's not in our metric space at all. Then, does $B$ contain all its limit points?
Alternatively, we can say $B$ every point in $B$ is an interior point (has a neighbourhood contained in $B$) so that makes $B$ open.
So then is $B$ both open and closed?
Thanks
Best Answer
You must have proved that a set is open in the induced metric iff it is the intersection of an open set with the subspace, and the analogous statement holds for closed sets (note that this is the definition of induced topology, if you know topology).
Since we have that $B=[-\sqrt2, \sqrt 2] \cap \mathbb{Q}=(-\sqrt{2}, \sqrt{2}) \cap \mathbb{Q}$, it follows that $B$ is closed (by the first equality) and open (by the last).