Compactness of a Set Course Title : Real Analysis 1 Course Code : MTH-211 Submitted to Submitted by Abdullah Al Mahbub Maruf Ahmed Abeer Associate Professor Id: 12004025 Department of Mathematics Session: 2019-20 Comilla University Department of Mathematics Comilla University Sneak peek ▪ Open cover of a set ▪ Subcover of a set ▪ Compact Set Compactness Open Cover of Set Let 𝑆⊂ R and 𝐶 = {𝐴 |𝑖 ∈ 𝐼} be a collection of open sets then C is said to be open cover of S if 𝑆 𝑖" ∈" 𝐼 𝐴𝑖 Example : Let S = R (Real numbers) and C = {(-n, n) | n ∈ N} then 𝑺 ⊂ ( …..-4 ( -3 ( -2 ( -1 ) 1 𝒏∈𝑵 (−𝒏, 𝒏) ) 2 ) 3 ) 4 ……. C = {(-n, n) | n ∈ N} is an open cover of R Subcover: A sub-collection of open cover of any set is called a subcover. Example : C = {(-n, n) | n ∈ N} and D = {(-3n, 3n) | n<4 & n ∈ N} Here both are open cover of R but D is subcover of C Compact Set: If a finite subcover of S is contained in each open cover of any set S, then S is called a compact set. Example : A finite subset of real number is compact. Proof Let S = {a₁, a,....... a} is a finite set and G be the family of sub cover of S. | | a1 a2 | | a3… … … … … .. an Then ∃ a open set containing each a ∈ S (| ) (| ) (| ) ( | ) a1. a2 a3. ….. … … an Hence S is finite → open sets containing a are finite. →G has finite subcover of S ⇒ S is compact. Example – 2: R = {set of real number} is not compact. Proof: We know that R has a open cover C = {(-n, n): n ∈ N} and 'C' has infinite subcover like ((-kn, kn): k N, n ∈N} → R is not compact. The End