# Prove that $\Bbb Q$ is dense in $\Bbb R$ constructed by Cauchy sequences

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 $$[(a_n)] \prec [(x_n)] \prec [(b_n)]$$.

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

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

• 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? – Arthur Feb 10 at 10:50
• 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. – Henno Brandsma Feb 10 at 11:11
• @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. – Henno Brandsma Feb 10 at 11:16
• nitpick: the $\varepsilon$ must be chosen rational or even of the form $\frac{1}{n}$ with $n>0$ a natural number. – Henno Brandsma Feb 10 at 11:29
• 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. – Henno Brandsma Feb 10 at 11:57