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Understanding Compact Spaces in Metric Spaces

In metric spaces, compact sets play a crucial role in analysis. A set ( A subseteq X ) is compact if every open cover has a finite subcover. In finite sets, compactness is straightforward as every finite set is compact. This guide explores various definitions, theorems, and examples, including proofs of compactness for specific sets in the standard metric space ( (R,d) ). It also highlights the famous Heine-Borel theorem, connecting compactness with closed and bounded sets in Euclidean spaces.

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Understanding Compact Spaces in Metric Spaces

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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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