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Math Help - Cauchy equivalence relation question

  1. #1
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    Cauchy equivalence relation question

    Show that if \{a_n\} and \{b_n\} are Cauchy in \mathbb{Q} contained in equivalence classes a and b (real numbers) respectively, and if a_n < b_n for all n, then there exists a real number \delta which contains a rational Cauchy sequence \{\delta_n\} each of whose \delta_n is positive and b=a+\delta.

    Here's what I have, for some reason it doesn't seem correct.

    since a_n < b_n then 0 < b_n-a_n implying that there exist \delta_n such that \delta_n = b_n-a_n for all n but since {\delta_n} is an equivalence class of a Cauchy sequence then [\{\delta_n\}] = \delta \in \mathbb{R} and the same for [\{a_n}]=a \in \mathbb{R} and [\{b_n}] = b \in \mathbb{R} so then [\{\delta_n\}] = [\{b_n\}]-[\{a_n\}]\Rightarrow \delta = b-a \Rightarrow a+\delta = b
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  2. #2
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    Quote Originally Posted by lllll View Post
    Show that if \{a_n\} and \{b_n\} are Cauchy in \mathbb{Q} contained in equivalence classes a and b (real numbers) respectively, and if a_n < b_n for all n, then there exists a real number \delta which contains a rational Cauchy sequence \{\delta_n\} each of whose \delta_n is positive and b=a+\delta.

    Here's what I have, for some reason it doesn't seem correct.

    since a_n < b_n then 0 < b_n-a_n implying that there exist \delta_n such that \delta_n = b_n-a_n for all n
    Be careful here. a_n and b_n are rational numbers of sequences in equivalence classes. You don't want to say that b_n- a_n is an equivalence class. \{\delta_n\} is a sequence of rational numbers and will be in an equivalence class (and so define an equivalence class) if you can prove it is a Cauchy sequence.

    but since {\delta_n} is an equivalence class of a Cauchy sequence then [\{\delta_n\}] = \delta \in \mathbb{R} and the same for [\{a_n}]=a \in \mathbb{R} and [\{b_n}] = b \in \mathbb{R} so then [\{\delta_n\}] = [\{b_n\}]-[\{a_n\}]\Rightarrow \delta = b-a \Rightarrow a+\delta = b
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