Understanding compact subsets of metric spaces

Let $(X,d)$ be a metric space, a subset $S \in X$ is called compact, if any infinite sequence $\{x_{n}\}_{n\in\Bbb N}\in S$ has a sub-sequence with a limit in S.

1. What does "if any infinite sequence" mean? Maybe: At least one, all?

2. What does "has a sub-sequence" mean? Maybe: At least one? Exactly one?

I am no mathematician and I don't understand the (practical) relevance of this property. Please explain it to me.

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Jede Folge aus $S$ besitzt (wenigstens) eine konvergente Teilfolge, deren Grenzwert in $S$ ist. – Brian M. Scott Jul 14 '12 at 20:52
This is one of those ambiguous uses of the word "any". "...if any infinite sequence has..." could reasonably mean "...if there is any infinite sequence that has...", so that "any" in effect is the same as "some". But that is not what is meant. It could also reasonably mean "...if it is the case that any sequence (no matter which one) has...". That makes it in effect the same as "every". So I would just write "every". – Michael Hardy Aug 17 '12 at 22:57

The definition can be reformulated as follows - "Let $(X,d)$ be a metric space, a subset $S∈X$ is called compact, if for all infinte sequences $\{ x_{n}\}_{n=1}^{∞}\subseteq S$ the following holds: $\{ x_{n}\}_{n=1}^{∞}$ has a concentration point and if $\bar{x}$ is a concentration point of $\{x_{n}\}_{n=1}^{∞}$ , than $\bar{x} \in S$." Now consider the definition of a concentration point of a infinite sequence. The practical side of compactness of a given set is that it contains it's "edge". If any definition seem vague to you, try to rewrite it using relevant notions with which you're more familiar with.

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Sorry, all my previous math education has been in german: concentration point = convergence point = limit? – tauran Jul 14 '12 at 11:51
No, concentration point = Häufungspunkt. Sorry for confusion, according to approved terminlology, it should stand accumulation point instead of concentration point. – data Jul 14 '12 at 11:58
I think I got it. It's easy to confuse limit point (or accumulation point) of a set and limit of a sequence :-) – tauran Jul 14 '12 at 12:23

The definition that you quoted is in fact that of sequential compactness. A subset $K$ of a topological space $X$ is called compact (see I J Maddox, p.62) if any open cover has a finite sub cover. Precisely, if $\{G_{\alpha}\}$ is a collection of open sets that covers $K$, then there exists a finite collection $G_{\alpha_1},G_{\alpha_2},...,G_{\alpha_n}$ which covers $K$.

Now a $metric$ space is said to be sequentially compact if and only if every sequence has a convergent subsequence. There is a theorem that establishes a link between the two: a metric space is compact if and only if it is sequentially compact (Maddox, Theorerm 21).

1. Any sequence means any sequence, It is not sufficient to prove this for one particular sequence.

2."Has a sub-sequence" means "at least one". Once you have found one, you can extract a convergent sub-sub-sequence from it.

Compact spaces are useful, because they allow one to construct convergent sequences which are themselves of paramount importance in proving many results.

For example, there is a theorem in approximation theory (see M J D Powell, Theorem 1.1.) that says that asserts the existence of the best approximation to an element of a linear space from a compact subspace.

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Compactness is a subtle "finiteness" property. Unfortunately there is no simple way to characterize resp. to define it in general. Only for subsets $S\subset {\mathbb R}^n$ there is a simple characterization: $S$ has to be closed and bounded. Bounded means, of course, that $S$ should fit in a ball of finite radius, while closed means that the boundary $\partial S$ is included in $S$.

Now the practical consequences of being compact are huge. The simplest consequence is that a a continuous function $f:\ S\to{\mathbb R}$ on a compact set $S$ is automatically bounded, i.e., there is an $M$ such that $|f(x)|\leq M$ for all $x\in S$. But more is true: You are guaranteed at least one point $\xi\in S$ where $f$ takes its global maximum: There is a $\xi\in S$ with $f(\xi)=\max_{x\in S}f(x)$.

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