Consider the following subset of $\mathbb{R}^{2}.$ The set of all complex numbers (i.e. $\mathbb{R}^{2}$).
I'm trying to show that this set is Closed, Open, Perfect but not Bounded.
Closed:
let us denote the set of all complex numbers by $E$. Then $E^{c} = \emptyset.$ It's vacuously true that $\emptyset$ is open. Hence $E$ is closed.
Open:
Similar to the above argument. The empty set is also Closed. Hence $E$ is open.
Perfect:
We need only, check that every point is a limit point. Let $z \in E.$ Then for $\epsilon > 0$ consider $B_{\epsilon}(z) = \{w: d(w,z) < \epsilon\}.$ I know that this is nonempty and therefore every point is a limit point, but I'm not really seeing how to explain why. (In my mind, I'm just thinking of drawing a small circle around the point $z$ then since we are dealing with Complex numbers it will of course contain another point!)
Not Bounded:
We need to show that for any $M \in \mathbb{R}$ and $z \in \mathbb{R}^{2}$, there is some $w \in \mathbb{C} \subset \mathbb{R}^{2}$ such that $d(z,w) > M.$
I'm not too sure how to mathematically show it, but the fact it's the entire complex plane, it is clearly going to not be bounded! Is there a general strategy for proving bounded / not bounded properties of sets? This seems to be the part I struggle on most.
Show that a subset is bounded if it is contained in some open/closed ball. Clearly $\mathbb{C}$ is not contained in an open bal, so $\mathbb{C}$ is not bounded.