# Intuition behind compact subspaces of a metric space

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