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I've read up on compactness in a metric space and have found a few definitions (let $X$ be a metric space and $E \subset X$ in all the following):

  1. $E$ is compact in $X$ if for every open covering of $E$ in $X$, $\{G_{\alpha \in A}\}$, $\exists$ a finite sub cover, $\{G_{\alpha \in B}\}$ s.t. $E \subset \cup_{\alpha \in B} G_{\alpha}$
  2. $E$ is compact if every sequence in $E$ has a convergent subsequence.
  3. $E$ is compact in $X$ if it is closed and bounded in $X$.

Now, I like the last definition better because it's easiest to understand, but I have a sneaking feeling they all are equivalent statements. From the first definition I can gather that the third definition follows, but I can't seem to wrap my mind on the idea that if $E$ is bounded and closed in $X$ then it has a finite subcovering in $X$. Then the second definition I can't really understand at all... I can't seem to intuitively relate it to either of the other definitions. I suppose my confusion stems from which of these definitions is best for describing intuitively what a compact subspace is?

EDIT: Changed the 2nd definition as the first comment points out that the original one was wrong

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I'm not sure your second statement is true. I may be missing something, but if we let $E = [0,1)$ in $\mathbb{R}$ with the usual metric, $E$ is not compact, and yet the set of limit points is $[0,1] \not=\{\}$ (assuming you mean that $\{\}$ is the empty set)? Could someone clear this up for me? I'm used to the first and third definitions of compactness, but normally also see the statement that "$E$ is compact if every sequence in $E$ has a convergent subsequence". – Tom Oldfield Apr 7 '13 at 17:59
Ah yea that would make more sense - I'll change the question to fix the 2nd definition – DanZimm Apr 7 '13 at 18:05
Your last assertion is not true. If you take $X$ to be an infinite set with the discrete metric, ie. $d(a,b) = 1$ for $a \neq b$, then $X \subseteq X$ is a closed and bounded subset of itself. But it is not compact. – Piotr Pstrągowski Apr 7 '13 at 18:10
The least assertion is not true in general metric spaces, but rather, true in $\Bbb R^n$ with the usual topology. – Pedro Tamaroff Apr 7 '13 at 18:11
However, the statement 3. is true in the very special case of $\mathbb{R}^{n}$ with the usual metric, this fact is known as a Heine-Borel theorem. – Piotr Pstrągowski Apr 7 '13 at 18:11
up vote 4 down vote accepted

Your first definition is the standard one (both for metric spaces and topological spaces in general). Your third definition is true in Euclidean space (it's the Heine-Borel theorem), but not every metric space. An arbitrary metric space is sometimes defined to be compact if it is totally bounded and complete. Your second definition can be proved from this one. Note that definitions one and two are only equivalent in metric spaces (not in general topological spaces).

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Ok, so, intuitively what is a compact metric space? – DanZimm Apr 7 '13 at 18:58
It's a difficult concept to grasp intuitively. I've heard it described as way to give infinite spaces a finite-like characteristic. There are some good answers here (better than I could give). – manthanomen Apr 7 '13 at 19:12
Did not see that thread before I asked this, thank you! – DanZimm Apr 7 '13 at 23:07

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