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Compact Spaces: Definition:

Compact Spaces: Definition: Let (x,d) be a metric space and A  X , {G i I } be a family of open sets such that A . then we say that {G i  I } be an open cover of A. If I is a finite set then we that {Gi  I } a finite open cover of A. Definition:

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Compact Spaces: Definition:

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  1. Compact Spaces: Definition: Let (x,d) be a metric space and AX , {GiI} be a family of open sets such thatA then we say that {GiI} be an open cover of A. If I is a finite set then we that {GiI}a finite open cover of A. Definition: Let (X,d) be a metric space and then we say that A is a compact set in X if for every {GiI} be an open cover of A contains a finite open cover of {GiI} of A. if A=X the we say that (X,d) be a compact space. Example: Let (R,d) be an usual metric space Prove that {1}R be a compact set. Proof: Let {UiI} be an open cover of {1} , then {1} and so there exists an open set Uk such that {1}Uk . Hence {Uk} be a finite open cover of {1}. Therefore {1} is compact set.

  2. Example: Let (X,d) be a metric space . Prove every finite set in X is compact set. Proof: Let A be a finite set, then A={x1 , x2 , x3 , … , xn } To prove that A is a compact set. Let {Gi: i} be an open cover of the set A, then x1  G1 such that x1G1 x2  G2 such that x2G2 xn  Gn such that xnGn Thus Hence {Gi: i=1,…,n} is a finite open cover of A. Therefore A is a compact set

  3. Example: Prove that the usual space (R, d) is not compact space. Proof: For prove that R is not compact set , we need to prove that (there exists an open cover of R such that this cover has not finite open cover of R). Consider this family : {-k,k : k=1,2,3, … } -for each n the interval -k,k is open set -By Archemed’s theorem, for all xR there exist kZ+ , such that k > x -k < x < k x -k,k Thus Hence the family {-k,k : k=1,2,3, … } is an open cover of R . But this cover has not finite cover of R.

  4. If it has finite cover of R, and suppose {-ki,ki : k=1,2,3, … , n} Therefore , let r=max{k1,k2,…,kn} We not that rR but r]-n,n[ And this contradiction. Therefore R is not compact set. Theorem: Let (X,d) be a metric space and A be closed set in X, then A is a compact set in X. Proof: Since A is closed then Ac is open set. Let {Gi: i} be an open cover of A, then We note that X=AAc  Ac

  5. Thus we have a new open cover {GiAc : i} of X. Since X is compact set then there exists 1,2,…,n such that Now Thus Hence {Gi:i=1,2,…,n} is a finite open cover of A. Therefore A is a compact set.

  6. Theorem(Hein-Borel): Let (R,d) be the usual metric space and AX. A is a compact set in X iff A is closed and bounded set.

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