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**Unformatted text preview: **MATH 5333: Assignment #3 Section 14 3. (a) Here are two open covers of [1 , 3) which do not have finite subcovers. G = braceleftbig( 1- 1 n , 3- 1 n )bracerightbig , n = 1 , 2 , 3 , . .. } F = braceleftbig(- 2 , 3- 1 n )bracerightbig , n = 1 , 2 , 3 , . . . } (b) Here are two open covers of N which do not have finite subcovers. G = { (0 , n ) , n = 1 , 2 , 3 , . . . } F = { ( n- 1 / 2 , n + 1 / 2 , n = 1 , 2 , 3 , . .. } 4. Let } be a collection of compact sets and let C = intersectiondisplay B ∈G B . Choose an element B ∈ G . Then C ⊆ B . Since B is bounded, C is bounded. C c = parenleftBigg intersectiondisplay B ∈G B parenrightBigg c = uniondisplay B c . Since each B ∈ G is compact, each B is closed, and so each B c is open. The union of open sets is open. Therefore, uniondisplay B c is open and C = intersectiondisplay B ∈G B is closed. Thus, C is closed and bounded, which implies that C is compact. 5. (a) If S 1 and S 2 are compact sets, then S 1 and S 2 are closed and bounded. Let T = S 1 ∪ S 2 . Then T is closed since the union of any collection of closed sets is closed. Since S 1 is bounded, there is a positive number M 1 such that | s | ≤ M 1 for all s ∈ S 1 . Similarly, there is a positive number M 2 such that | s | ≤ M 2 for all s ∈ S 2 . Let M = max { M 1 , M 2 } . Choose any s ∈ T . Then either s ∈ S 1 or s ∈ S 2 . In either case, | s | ≤ M . Therefore T is bounded. T is compact since it is closed and bounded. 8. (a) Let G be an open cover of T . Then F = G ∪ T floorright is an open cover of S . Since S is compact, F has a finite subcover { G 1 , G 2 , . . ., G n , T c } . The sets { G 1 , G 2 , . . ., G n } form a finite subcover of T ....

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- Math, Topology, 1 k, Metric space, 1 k, 1 5 2k