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Math Help - proof that a empty set is bounded

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    proof that a empty set is bounded

    prove/ disprove that every non empty subset of Q that is bounded above has a supremum in Q
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    Quote Originally Posted by jmheaney View Post
    prove/ disprove that every non empty subset of Q that is bounded above has a supremum in Q
    Consider the set: \left\{ {x \in \mathbb{Q}^ +  :x^2  < 2} \right\}.
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    Quote Originally Posted by jmheaney View Post
    prove/ disprove that every non empty subset of Q that is bounded above has a supremum in Q
    ??? You titled this "proof that an empty set is bounded"! This has nothing to do with the empty set!

    Since that is false, a counter-example suffices. Let A= {3, 3.1, 3.14, 3.141, 3.1415, 3.14159, ...} where each number in the sequence is \pi to one more decimal place. Since the real numbers are "complete", this statement is true for R- there is a supremum in R. If this set had a supremum in Q, it would have to be the same as the supremum in R. What is the supremum of this set (in R). Is that a rational number?
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