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Let $CS(\mathbb{Q})$ denote the set of all Cauchy sequences of rational numbers. Define an equivalence relation $\sim$ on $CS(\mathbb{Q})$ as follows: $$(x_n)\sim (y_n)$$ if $\lim_{n\to \infty}(x_n-y_n)=0$. It follows that $\mathbb{Q}\subset CS(\mathbb{Q})/\sim,$ and the claim is that $\mathbb{R}=CS(\mathbb{Q})/\sim$.

I can define $+,\cdot,\le$ on $CS(\mathbb{Q})/\sim$, I can prove that it is a field, and I can prove that it has the least upper bound property. However, how am I sure that this is enough to claim that the resulting field is equivalent to $\mathbb{R}$?

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    $\begingroup$ what is your definition of the field $\mathbb R$? $\endgroup$ – Ittay Weiss Aug 17 '14 at 8:19
  • $\begingroup$ $\mathbb{R}$ is the smallest set containing $\mathbb{Q}$ which has the least upper bound property. $\endgroup$ – The Substitute Aug 17 '14 at 8:21
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If $F$ is an arbitrary ordered field with the least upper bound property, then it must contain (as a field) two elements you'll usually call 0 and 1, so it must contain $1+1$ which you'll call $2$, etc. From this you can show that $F$ must contain an isomorphic copy of $\mathbb Q$ and now you use the upper bound property to find an isomorphism from $F$ to $\mathbb R$.

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Every irrational number has a corresponding Cauchy sequence associated with it. For example, $\pi$ has $3,3.1,3.14,\dots$

Similarly, every rational number also has a corresponding Cauchy sequence associated with it. For example, $2$ has the Cauchy sequence $2,2,2\dots$

I'm sure you can now develop a proof.

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