If A is in B then sup(A) < sup(B)

  • #1
Mr Davis 97
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Homework Statement


Let ##X## and ##Y## be nonempty subsets of real numbers such that ##X \subseteq Y## and ##Y## is bounded above. Prove that ##\sup X \le \sup Y##

Homework Equations




The Attempt at a Solution


Case 1: ##X = Y##. Trivially, ##\sup X \le \sup Y##.

Case 2: ##X \subset Y##. Then there exists a ##y^* \in Y## that is an upper bound for ##X##. Since ##\sup X## is the least upper bound, ##\sup X \le y^*##. But ##y^* \le \sup Y## since ##\sup Y## is an upper bound for ##Y##. So ##\sup X \le y^* \le \sup Y##, and so ##\sup X \le \sup Y##
 

Answers and Replies

  • #2
fresh_42
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Why should ##y^*## (= any upper bound) be less than ##\operatorname{sup}Y## (= least upper bound)? The contrary seems right, especially since I cannot see any reason why ##y^*\in Y##. However, you don't need ##y^*## at all. Why not work directly with ##\operatorname{sup}Y## as upper bound for both?
 
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