What is a compact set?

Question

According to one of the definition I found:

A set $S \subset R^n$ is compact if every sequence in $S$ has a convergent subsequence, whose limit lies in $S$.

Specifically, I am not clear what does convergent subsequence mean but it would be really helpful if someone can explain what's a compact set by breaking down the above definition into sub parts. Also, an example and a counter-example will be really helpful.

This definition has been used in the context of game theory to proof existence of Nash equilibrium.

Answer 1

Let's take $[0,1]\subseteq \Bbb R$ as an example of a compact set. There are many sequences contained in that set. Some converge, others do not. For instance, $1, \frac12, \frac13,\ldots$ does converge, while $0,1,0,1,0,\ldots$ does not.

However, if you look at that second sequence, we can "remove" every other term, and be left with $0,0,0,\ldots$, which is a converging sequence. That's what a "convergent subsequence" is. "Convergent" because it, well, converges. And "subsequence" because it's a sequence made of terms from the original sequence. (Technical note: in a subsequence, the terms must appear in the same order as they did in the original sequence, even though not all terms are present. You are allowed to remove terms, but not shuffle them or duplicate them when you make a subsequence.)

If a sequence is convergent, then all (infinite) subsequences will necessarily converge as well, to the same limit.

So $[0,1]$ being compact means that for any sequence $a_n$ with $a_n\in [0,1]$, it is possible to remove terms from that sequence to make a subsequence which is convergent, and where the limit is contained in $[0,1]$. That last part is important, because $(0,1)$ isn't compact (or, rather, we don't want it to be, so we make a definition which excludes it). This is illustrated by the sequence $1, \frac12, \frac13,\ldots$ which converges to $0$ (and so does any subsequence). But $0$ is not contained in $(0,1)$, so in this case we have found a sequence in which all convergent subsequences converge to a limit outside $(0,1)$.

Answer 2

In other terms, the definition says that any infinite sequence of points has an accumulation point (points getting arbitrarily close to a given one), and that the accumulation point belongs to the set.

This property could be invalidated in two ways:

• by having points escaping to infinity (then no accumulation point),

• by having an accumulation point in the closure of the set but not in the set itself.

Compactness, i.e. being bounded and closed ensures that this does not happen. This is useful when a proof requires that a limit point lies in the set.