The two definitions of a compact set 
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* In general, $A$ is compact if every open cover of $A$ contains a finite subcover of $A$.

* In $R$, $A$ is compact if it is closed and bounded.

The second is very easy to understand because I can easily come up with an example like $[0,1]$ which is both closed and bounded so it's compact.
However, I am very confused at definition (1) because I don't really understand what is meant by a cover and I don't understand how this is really related to a set being closed and bounded?
Could someone please explain what is the relationship between (1) and (2)?
Thank you.

The definitions of open cover and finite subcover can be stated simply.  An open cover of $A$ is a collection of open sets in the underlying topology whose union includes all of $A$.  This definition of compactess says that if you have any open cover of any set $A$, you should be able to find a finite collection of sets in that cover that also cover $A$.  As an example, let $A=[0,1]$ and my collection of sets be $(\frac 1{2n-1},2), n \in \Bbb N \cup (\frac {-1}{1000},\frac 1{1000})$.  Every point in $[0,1]$ is in at least one of these sets, so it is a cover.  We can find a finite subcollection by requiring that $n \lt 1000$ in the first piece that also covers $[0,1]$.  To prove that $[0,1]$ is compact under this definition you would have to show that finding such a finite subcover was possible for any cover.  On the other hand, let $B=\Bbb R$.  We can cover $B$ with $(2n, 2n+2), n \in \Bbb Z$ but no finite subcollection covers $\Bbb R$, so $\Bbb R$ is not compact. The reason we specify an open cover is that we could cover $[0,1]$ with $[\frac 1{n+1}, \frac 1n] \cup \{0\}$ and not have a finite subcover. We want $[0,1]$ to be compact, so need to choose a definition that makes it so.
