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# Category Archives: Compact

## Winter 2008 #5

Problem Statement: Let . Prove is Riemann Integrable on and . Proof: First note that each is continuous on and since is a sum of continuous functions it follows that is continuous on . Furthermore, since is a compact set it … Continue reading

## Spring 2004 #5

Problem Statement: Let be differentiable. Let such that and . If is between and then there exists such that . Proof: Wlog assume . Fix . Define . Then . Note that and since . and so for sufficiently close to , … Continue reading

Posted in Analysis, Compact, Continuity, Differentiable, Math
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## Spring 2008 #1

Problem Statement: Let be a compact subset of and a continuous function. Show there exists a such that for every . Proof: Since is compact and is continuous it follows that is compact. Bolzano-Weierstrass gives us that is closed and bounded. This means … Continue reading

Posted in Analysis, Compact, Continuity, Math
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## September 2000

Problem Statement: Use the open cover definition of compactness to show that any finite union of compact sets is compact. Proof: Let be compact sets in . We wish to show that is a compact set. Let be an open cover of … Continue reading

Posted in Analysis, Compact, Topology
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## May 1990 #4

Problem Statement: Using properties of the Riemann Integral show that if is Riemann Integrable on and then is uniformly continuous on . Proof: We wish to show that for every there exists a such that for every in it follows that . … Continue reading

Posted in Analysis, Compact, Uniform Continuity
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## Winter 1995 #1

Problem Statement: Using only the definition of a Cauchy sequence show that every Cauchy sequence in is bounded. Proof: Let be a Cauchy sequence in . Let , then there exists a such that for every it follows that . Let latex … Continue reading

Posted in Analysis, Compact, Math, Sequence
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## September 2001 #5

Problem Statement: Use the open cover definition of compact to show that is a compact subset of . Proof: Let be an open cover of , for some indexing set . If is a finite cover then we’re done, so suppose that … Continue reading