'''Definition''' A set ''E'' is ''compact'' if and only if, for every family 
$$\{G_{ \alpha } \}_{\alpha \in A}$$ of open sets such that 
$$E \subset \bigcup_{\alpha \in A}G_{\alpha}$$, there is a finite set
$$\{\alpha_1 ,..., \alpha_n \} \subset A$$ such that 
$$E \subset \bigcup_{i=1}^{n} G_{\alpha_i}$$.

'''Example''': Let ''E''=(0,1] and for each positive integer ''n'', let 
$$G_n = \left(\frac{1}{n},2\right)$$. If 
$$0<x \leq 1$$, there is a positive integer n such that 
$$\frac{1}{n} < x$$; hence, 
$$x \in G_n$$, and thus 

$$$E \subset \bigcup_{n=1}^{\infty}G_n$$$ 

If we choose a finite set 
$$n_1,...,n_r$$ of positive integers, then

$$$\bigcup_{i=1}^{r} G_{n_i}=G_{n_0}$$$

where 
$$n_0=\max\{n_1,...,n_r\}$$ and 

$$$E \not\subset G_{n_0}=\left(\frac{1}{n_0},2\right)$$$

Thus, we have a family of open sets 
$$\{G_n\}_{n \in J}$$ such that 
$$E \subset \bigcup_{n \in J} G_n$$, but no finite subfamily has this property. From the definition, it is clear that ''E'' is not compact.

'''Heine-Borel Theorom''': A set $$E \subset \mathbb{R}$$ is compact iff $$E$$ is closed and bounded.

'''Examples:''' 

   * [2,8] is a compact set. 
   * The unit disk including the boundary is a compact set. 
   * (3,5] is not a compact set. 

Note that all of these examples are of sets that are uncountably infinite. 

''Definitions from: Introduction to Analysis 5th edition by Edward D. Gaughan''

'''Theorom:''' The union of compact sets is compact.