The definition that I have for a compact set is that "If a set of complex numbers S is compact then every sequence in S has a point of accumulation in S". For clarity here is the

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.

  • $\begingroup$ No, a sequence $\langle z_n:n\in\Bbb N\rangle$ in a compact set does not necessarily converge. It does necessarily have an accumulation point $w$, meaning that for each $\epsilon>0$ and $n\in\Bbb N$ there is a $k\ge n$ such that $|z_k-w|<\epsilon$. Your first question isn’t clear: all points of what? $\endgroup$ Jul 3, 2020 at 16:21
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    $\begingroup$ I like to think of compact sets as "approximately finite". Finite sets have the property that, if I take a sequence of points taking only finitely many values, I'll have to revisit one of the points infinitely many times. Another way of thinking about this is that, each sequence on a finite set must have a constant subsequence. Compact sets are a generalisation of that: sequences may not have constant subsequences, but they do have convergent subsequences, which are the next best thing. These subsequences converge to accumulation points of the sequence. $\endgroup$
    – user804886
    Jul 3, 2020 at 16:26
  • $\begingroup$ @EricTowers Sorry I forgot to put that in. I just made sup imaginary complex set. I now put it in the post as "radius of 1" $\endgroup$ Jul 3, 2020 at 16:34

2 Answers 2

  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.

  1. 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).

  • $\begingroup$ Thank you very much for your answer. It really gave me a very visual an deep understanding of what a compact set is. I just have one question, on the page following the definition of a compact set Lang proves that a compact set is one that is closed and bounded. I understood completely his proof for the bounded bit but the part where he proves the set is closed is unclear to me. My question is, is the set closed because if it was to be open then we would not be able to partition it in the way you have described above, and furthermore, there won't be a point of accumulation =>the set is closed? $\endgroup$ Jul 4, 2020 at 16:53
  • $\begingroup$ @MathsWizzard : You may notice I was ambiguous whether I was talking about the open disk, closed disk, or unit disk containing some subset of its boundary. This is because all of those are precompact sets. But let us observe that the open unit disk is not compact: it is easy to make a sequence, converging in the plane, that does not converge in the open disk. Here's one: $(1 - 1/n)_n$. So the open disk is not compact -- it contains sequences converging to points outside the open disk. $\endgroup$ Jul 6, 2020 at 15:41

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 $S$.

  • $\begingroup$ Hi thanks for the response. You are right, I had the definition the wrong way around, I have fixed it now by taking a screenshot of the definition in the book. Regarding the point of accumulation, what I don't understand is how could one interpret a point of accumulation for a random sequence of elements? $\endgroup$ Jul 3, 2020 at 16:51
  • $\begingroup$ On another note, my book defines a point of accumulation of an infinite set S as a complex number v such that given an open set U containing v, there exist infinitely many elements of S in U; which got me thinking, if I can have all of my points as points of accumulation can't I unite all of the open sets around each point of accumulation to show that S is open which is a contradiction as compact sets are closed and bounded? $\endgroup$ Jul 3, 2020 at 16:51
  • $\begingroup$ If $S=\{0\}$, then $S$ is compact. But the union of all open sets to which $0$ belongs is $\Bbb C$, which is not compact. $\endgroup$ Jul 3, 2020 at 16:53

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