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Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!


Let $\mathcal{C}$ be the set of Cauchy sequences of rationals. We define an equivalence relation $\sim$ on $\mathcal{C}$ by $$(a_n) \sim (b_n) \iff \forall \epsilon >0, \exists N, \forall n>N: |a_n - b_n| < \epsilon$$

Let $\mathcal{C} / {\sim}$ be the set of all equivalence classes of Cauchy sequences of rationals. We define a relation $\preccurlyeq$ on $\mathcal{C} / {\sim}$ by $$[(a_n)] \preccurlyeq [(b_n)] \iff \forall \epsilon >0, \exists N, \forall n>N: a_n - b_n < \epsilon$$

I have shown that $\preccurlyeq$ is a complete linear ordering here.

I have shown that $[(a_n)] \prec [(b_n)] \iff \exists \epsilon >0, \exists N, \forall n>N: \epsilon \le b_n -a_n$ here.

Theorem: $\Bbb Q$ is dense in $\Bbb R$.


My attempt:

$[(a_n)] \prec [(b_n)] \implies \exists \epsilon' >0, \exists N', \forall n>N': \epsilon \le b_n -a_n$.

$(a_n),(b_n)$ are Cauchy sequences $\implies$ $\exists N'', \forall m,n > N'': |a_m - a_n| < \dfrac{\epsilon'}{4} \wedge |b_m - b_n| < \dfrac{\epsilon'}{4}$.

Let $N_0 = \max \{N'+1,N''+1\}$. It follows that $\epsilon \le b_{N_0} -a_{N_0}$ and $\forall n > N_0: |a_n - a_{N_0}| < \dfrac{\epsilon'}{4} \wedge |b_n - b_{N_0}| < \dfrac{\epsilon'}{4}$.

Let $(x_n)$ be the sequence whose all elements are equal to $\dfrac{a_{N_0} + b_{N_0}}{2}$. It follows directly that $[(x_n)]$ is rational.

Next we prove $[(a_n)] \prec [(x_n)] \prec [(b_n)]$.

$n > N_0 \implies |a_n - a_{N_0}| < \dfrac{\epsilon'}{4} \implies a_n < a_{N_0} + \dfrac{\epsilon'}{4} \implies$ $\dfrac{a_{N_0} + b_{N_0}}{2} - a_n > \dfrac{b_{N_0}-a_{N_0}}{2}-\dfrac{\epsilon'}{4} \ge \dfrac{\epsilon}{2}-\dfrac{\epsilon'}{4} = \dfrac{\epsilon'}{4} \implies$ $\exists \dfrac{\epsilon'}{4} >0, \exists N_0, \forall n>N_0: \dfrac{\epsilon'}{4} \le \dfrac{a_{N_0} + b_{N_0}}{2} -a_n \implies [(a_n)] \prec [(x_n)]$.

$n > N_0 \implies |b_n - b_{N_0}| < \dfrac{\epsilon'}{4} \implies b_{N_0} - \dfrac{\epsilon'}{4} < b_n \implies$ $b_n - \dfrac{a_{N_0} + b_{N_0}}{2} > \dfrac{b_{N_0}-a_{N_0}}{2}-\dfrac{\epsilon'}{4} \ge \dfrac{\epsilon}{2}-\dfrac{\epsilon'}{4} = \dfrac{\epsilon'}{4} \implies$ $\exists \dfrac{\epsilon'}{4} >0, \exists N_0, \forall n>N_0: \dfrac{\epsilon'}{4} \le b_n - \dfrac{a_{N_0} + b_{N_0}}{2} \implies [(x_n)] \prec [(b_n)]$.

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  • $\begingroup$ You are using "dense" in the order theoretic sense. Using it in the real analysis sense (any real has a sequence of rationals converging to that real) is much more natural to the Cauchy construction of the reals, and therefore much easier. Is there a specific reason you didn't do it that way, or did you just not consider it? $\endgroup$ – Arthur Feb 10 at 10:50
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    $\begingroup$ Maybe you should make it explicit that you identify $q \in \mathbb{Q}$ with the (class of the) constant sequence with value $q$ (which is indeed Cauchy), so that $\mathbb{Q} \subseteq \mathbb{R}$ makes sense. $\endgroup$ – Henno Brandsma Feb 10 at 11:11
  • $\begingroup$ @Arthur For that he has to define a topology on $\mathbb{R}$ (the set of classes) first and he cannot define a metric (circular, as a metric already needs the reals) so you have to use the order anyway. $\endgroup$ – Henno Brandsma Feb 10 at 11:16
  • $\begingroup$ nitpick: the $\varepsilon$ must be chosen rational or even of the form $\frac{1}{n}$ with $n>0$ a natural number. $\endgroup$ – Henno Brandsma Feb 10 at 11:29
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    $\begingroup$ In Hrbacek and Jech dense means order dense. Look it up in the index. It's defined twice (in my 2nd edition). So your proof indeed works. $\endgroup$ – Henno Brandsma Feb 10 at 11:57

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