'Compact'에 해당되는 글 2건

  1. 2018.05.02 compact set
  2. 2018.05.02 [optimization] Extreme Value Theorem

compact set

mathmatics 2018. 5. 2. 15:18

A subset S of a topological space X is compact 

if for every open cover of S, there exists a finite subcover of S.


즉, S의 open cover들에 대해서, S의 finite subcover(element가 유한한 open cover of S의 subset)가 존재하는 경우, S가 compact하다고 한다.


cover

A family gamma of nonempty subsets of X whose union contains the given set X (and which contains no duplicated subsets) is called a cover (or covering) of X


 For example of cover

there is only a single cover of {1}

namely {{1}}

However, there are five covers of {1,2}

namely {{1},{2}}{{1,2}}{{1},{1,2}}{{2},{1,2}}, and {{1},{2},{1,2}}.



minimal cover

minimal cover is a cover for which removal of one member destroys the covering property.


Open cover

A collection of open sets of a topological space whose union contains a given subset. 

Example of open cover

- an open cover of the real line, with respect to the Euclidean topology, is the set of all open intervals (-n,n), where n in N.
- The set of all intervals (1/n,1), where n in N\{0}, is an open cover of the open interval (0,1).


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Extreme Value Theorem


If a function f(x) is continuous on a closed interval [a,b], 

then f(x) has both a maximum and a minimum on interval [a,b]. 


If f(x) has an extremum on an open interval (a,b), 

then the extremum occurs at a critical point. 


This theorem is sometimes also called the Weierstrass extreme value theorem.


Proofs

The standard proof of the first proceeds by noting that f is the continuous image of a  compact set on the interval [a,b], 

so it must itself be compact.

Since [a,b] is compact, it follows that the image f([a,b]) must also be compact.


compact set on the interval [a,b] : finite open cover로 covering되는 set on the interval [a,b]

f는 continuous image이기 때문에 f 또한 compact 해야한다.

f 가 compact 하기 때문에, finite open cover가 존재 한다. 즉, extreme 값이 해당 domain에 대한 image set에서 존재한다.





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