Understanding Compact Sets (In Complex Analysis)

Question

The definition that I have for a compact set is that

The way that I interpret this is that I can pick any sequence and it should have a limit point (i.e. point of accumulation). However, I can't understand a couple of things about this definition.

1. Why is it not true that all points can be points of accumulation?

2. Also, if I pick the disc of radius 1 centred at the origin of the complex plane and make a sequence say $0.7 +0.31i$, $0.8$, $0.653 - 0.74i$, $0.92 + 0.01i$ and so on of random complex numbers (members of this closed disc that is compact) how does it nesseseraly converge?

If anybody could shine some light on this I would greatly appreciate it.

Answer 1

1. Any point in $S$ can be an accumulation point of a sequence in $S$. Let $p$ be a point in $S$, the sequence $(p)_{n \in \Bbb{N}}$ is a sequence in $S$, the constantly $p$ sequence, and it has one accumulation point, $p$.

It is worth remembering that a sequence can have more than one accumulation point. Consider the sequence

$$ \left( \begin{cases} 1 ,& n \text{ odd} \\ 0 ,& n \text{ even} \end{cases} \right)_{n \in \Bbb{N}} $$

It has both $0$ and $1$ as accumulation points, but converges to neither. Do not conflate accumulation point with convergent. It is the case that, for each accumulation point, there is a subsequence that converges to that accumulation point.

2. Your "random points" scheme works to prevent having a convergent. It does not prevent having a subsequence which converges to an accumulation point.

Suppose you didn't select your points randomly but instead chose your points to avoid having any accumulation point. Pick $\varepsilon > 0$. Each time you put down a point, you can only ever have finitely many in a disk of radius $\epsilon$ around any other point already in your sequence. So partition the unit disk into squares, with edges parallel to the coordinate axes at integer multiples of $\varepsilon/2$, including the upper and left edges and both upper vertices. (Why "${}/2$"? Because $\sqrt{2}/2 < 1$, so any radius $\varepsilon$ disk centered in that square covers that entire square.) (The orientation of the grid is entirely irrelevant -- just rotat the disk to any other orientation. The alignment of the disk is also irrelevant -- if you choose to offset the grid, you get the same result.) There are only finitely many squares, so there are only finitely many disks.

In each radius $\varepsilon$ disk, to prevent having a convergent, you are only permitted a finite number of points. But this means your sequence is only finitely long. So this can't happen, there is at least one disk that has infinitely many points in it. And this is true for any choice of $\varepsilon$, so an infinitely long sequence in the unit disk must have at least one accumulation point (and a subsequence which converges to that accumulation point).

Answer 2

That definition doesn't make sense. In order that the sentence “If a set of complex numbers $S$ is compact then every sequence in $S$ has a point of accumulation in $S$” makes sense, you must know beforehand what a compact set is.

However, it would be correct to define “compact set” as follows: a set $S\subset\Bbb C$ is compact if every sequence of elements has an accumulation point in $S$.

And, yes, if $S$ is compact, the every $p\in S$ can be an accumulation point of some sequence $(s_n)_{n\in\Bbb N}$ of elements of $S$; just take $s_n=p$ for each $n\in\Bbb N$.

And a random sequence of elements of a compact set $S$ doesn't have to converge. All that is required is that it has a subsequence that converges to an element of $S$.