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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. | '''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 $$E \subset \bigcup_{i=1}^{n} G_{\alpha_i}$$. |
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| 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. | '''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 |
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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. | $$$E \subset \bigcup_{n=1}^{\infty}G_n$$$ 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 $\usepackage{amsfonts} % $E \subset \mathbb{R}$$ is compact iff ''E'' is closed and bounded. '''Examples:''' * [2,8] is a compact set. * The unit disk including the boundary is a compact set. * (3,5] is not a compact set. Note that all of these examples are of sets that are uncountably infinite. ''Definitions from: Introduction to Analysis 5th edition by Edward D. Gaughan'' '''Theorom:''' The union of compact sets is compact. |
Definition A set E is compact if and only if, for every family $$\{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$$, but no finite subfamily has this property. From the definition, it is clear that E is not compact.
Heine-Borel Theorom: A set $\usepackage{amsfonts} % $E \subset \mathbb{R}$$ 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.