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| In general we simplify the definition to be: A compact set is a set which is closed (that is it contains its boundary points) and is bounded. | A set ''E'' is ''compact'' if and only if, for every family [[latex2($\{G_{ \alpha } \}_{\alpha \in A}$)]] of open sets such that [[latex2($E \subset \bigcup_{\alpha \in A}G_{\alpha}$)]], there is a finite set [[latex2($\{\alpha_1 ,..., \alpha_n \} \subset A$)]] such that [[latex2($E \subset \bigcup_{i=1}^{n} G_{\alpha_i}$)]]. '''Example''': Let ''E''=(0,1] and for each positive integer ''n'', let [[latex2($G_n = \left(\frac{1}{n},2\right)$)]]. If [[latex2($0<x \leq 1$)]], there is a positive integer n such that [[latex2($\frac{1}{n} < x$)]]; hence, [[latex2($x \in G_n$)]], and thus [[latex2($$E \subset \bigcup_{n=1}^{\infty}G_n$$)]] If we choose a finite set [[latex2($n_1,...,n_r$)]] of positive integers, then [[latex2($$\bigcup_{i=1}^{r} G_{n_i}=G_{n_0}$$)]] where [[latex2($n_0=\max\{n_1,...,n_r\}$)]] and [[latex2($$E \not\subset G_{n_0}=\left(\frac{1}{n_0},2\right)$$)]] Thus, we have a family of open sets [[latex2($\{G_n\}_{n \in J}$)]] such that [[latex2($E \subset \bigcup_{n \in J} G_n$)]], but no finite subfamily has this property. From the definition, it is clear that ''E'' is not compact. '''Heine-Borel Theorom''': A set [[latex2(\usepackage{amsfonts} % $E \subset \mathbb{R}$)]] is compact iff ''E'' is closed and bounded. |
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| Technical def: A set is said to be compact if for every open ["Cover"] there exists a finite SubCover which also covers the set. | ''From: Introduction to Analysis 5th edition by Edward D. Gaughan'' Also we have that the union of compact sets is compact. |
A set E is compact if and only if, for every family latex2($\{G_{ \alpha } \}_{\alpha \in A}$) of open sets such that latex2($E \subset \bigcup_{\alpha \in A}G_{\alpha}$), there is a finite set latex2($\{\alpha_1 ,..., \alpha_n \} \subset A$) such that latex2($E \subset \bigcup_{i=1}^{n} G_{\alpha_i}$).
Example: Let E=(0,1] and for each positive integer n, let latex2($G_n = \left(\frac{1}{n},2\right)$). If latex2($0<x \leq 1$), there is a positive integer n such that latex2($\frac{1}{n} < x$); hence, latex2($x \in G_n$), and thus
latex2($$E \subset \bigcup_{n=1}^{\infty}G_n$$)
If we choose a finite set latex2($n_1,...,n_r$) of positive integers, then
latex2($$\bigcup_{i=1}^{r} G_{n_i}=G_{n_0}$$)
where latex2($n_0=\max\{n_1,...,n_r\}$) and
latex2($$E \not\subset G_{n_0}=\left(\frac{1}{n_0},2\right)$$)
Thus, we have a family of open sets latex2($\{G_n\}_{n \in J}$) such that latex2($E \subset \bigcup_{n \in J} G_n$), but no finite subfamily has this property. From the definition, it is clear that E is not compact.
Heine-Borel Theorom: A set latex2(\usepackage{amsfonts} % $E \subset \mathbb{R}$) is compact iff E is closed and bounded.
Example [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.
From: Introduction to Analysis 5th edition by Edward D. Gaughan
Also we have that the union of compact sets is compact.