# Discrete and compact subset must be finite

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?

• Do you know the Heine-Borel Theorem. "Every bounded infinite set has a cluster point"? Commented Mar 9, 2014 at 1:04
• Nope, no idea what a cluster point is. Commented Mar 9, 2014 at 1:09
• There could be an issue with Terminology. Heard of a Limit Point? Accumulation Point? Commented Mar 9, 2014 at 1:17
• Acculumation point no, limit point yes but not in this class :( Commented Mar 9, 2014 at 1:17
• Discrete+closed $\nRightarrow$ finite: consider $\mathbb{Z}$. Commented Jun 7, 2016 at 5:11

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.

• Yep! I do know bolzano weierstrass. However I don't understand the last part of your proof. Can you explain exactly why $L$ being in $D$ and $D$ being discrete leads to the impossibility of a sequence that converges to $L$? Commented Mar 9, 2014 at 1:37
• I hope it is clear now, please accept my answer if it is, because i am new to stack exchange and i would like to build up my reputation :D Commented Mar 9, 2014 at 1:54
• it is clear, thank you. you mean "intersection" instead of union though! Commented Mar 9, 2014 at 1:58
• Yes indeed. I should probably use more LaTeX notation! Commented Mar 9, 2014 at 2:00

If $D$ is discrete, then for every $z\in D$, there exists a $r_z>0$, such that $B(z,r_z)\cap D=\{z\}$, where $B(z,r_z)$ is the disc centered at $z$ with radius $r_z$.

Let now the collection of open sets $${\mathcal C}=\{B(z,r_z): z\in D\}.$$ Then $\mathcal C$ is an open cover of $D$ without a proper subcover.

Since $D$ is compact, then the cover $\mathcal C$ is finite, and hence $D$ is finite.

• We haven't talked about open covers in class so I am not allowed to use that definition. It would make this easy, though. All I have is the definitions for closed and boundedness I listed above. Commented Mar 9, 2014 at 0:56

Your question here contains a perfect example of a bounded and discrete set in $\Bbb R$ which is not finite. You should be able to extend that to $\Bbb C$. I think it should be a good exercise for you to identify which one it is. And this question also contains a proof for a compact, discrete set to be finite. But I'm afraid it does use open covers. But that is the only one I know. Apologies for that.

And consider the set of all points in $\Bbb C$ which I believe you interpret to be the product set $\Bbb R \times \Bbb R$ which have integer coordinates. This set is closed, discrete but not finite.

So in conclusion, NO.

- A set that is discrete and bounded is not necessarily finite.
- A set that is discrete and closed is not necessarily finite.