Cauchy equivalence relation question

**lllll** · October 15th 2010, 10:57 PM

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 $[\{\delta_n\}] = [\{b_n\}]-[\{a_n\}]\Rightarrow \delta = b-a \Rightarrow a+\delta = b$

**HallsofIvy** · October 16th 2010, 05:20 AM

Originally Posted by lllll

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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