Show that a discrete and compact subset $D \subset \mathbb{C}$ must be finite. Does this conclusion hold if $D$ is just discrete and bounded? How about discrete and closed?
Compact is the usual (for these simple spaces), closed and bounded, where closed is contained under the limit operation/contains all limit points.
Bounded it can be contained in a ball of some radius around the origin.
$D \subset \mathbb{C}$ is a discrete subset if $\forall z \in D$ there exists a ball of radius $r>0$ such that $D \cap B_r(z)$ = $\{z\}$.
Okay, for bounded set: why is discrete required? Can you give mme an example of a bounded set that is NOT finite?
Best Answer
Proof without open covers:
Assume that $D$ is not finite, take an infinite sequence of distinct elements in $D$. The Bolzano-Weierstrass theorem (I hope you know this one) states that there is a subsequence , say $A_{n}$ that converges. But this is impossible, because a convergent sequence has a limit $L$ in $D$ (because $D$ is closed). But since $L\in D$ and $D$ is a discrete set, there is a $r \in \mathbb{R}$ so $B_{r}(L) \cap D = \{L\}$ This is in contradiction with the epsilon-delta definition of convergence. Take $\epsilon = r/2$ , then there is a $n \in \mathbb{N}$ such that norm of $(L-A_{i}) < r$ for each $i>n$
So D, can not be infinite.