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'''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''': 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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$$$E \subset \bigcup_{n=1}^{\infty}G_n$$$
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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. If we choose a finite set
$$n_1,...,n_r$$ of positive integers, then
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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. $$$\bigcup_{i=1}^{r} G_{n_i}=G_{n_0}$$$
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A set ''E'' is ''compact'' if and only if, for every family [latex2($\left{G_{\alpha}\right}_{\alpha \in A}$)] of open sets such that [latex2($E \subset \cup_{\alpha \in A}G_{\alpha}$)] where
$$n_0=\max\{n_1,...,n_r\}$$ and
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Introduction to Analysis 5th edition by Edward D. Gaughan $$$E \not\subset G_{n_0}=\left(\frac{1}{n_0},2\right)$$$

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.

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.

CompactSet (last edited 2020-01-26 17:51:19 by 68)