All real numbers can be expressed as a limit of rational numbers? RTP

Let $C$ be a set of Cauchy sequences. $\forall x \in {\Bbb R}, \exists \{a_n\} \in C$ sucht that ${a_n} \to x$.

I have no clue to even start this problem.
All I know so far is that $\Bbb R$ is a set of equivalent classes of the limit of rational Cauchy sequences. 
Can anyone give me a proof or a hint ?
I am new to analysis, so it would be nice if you could tell me the rationale of this problem as well.
 A: I’m going to write this answer on the assumption that you’re working through a fairly rigorous construction of the reals; if you’re not, it’s more technical than you actually need.
Let $x\in\Bbb R$; then by the definition with which you’re working, $x$ is an equivalence class of Cauchy sequences of rational numbers. Let $\sigma=\langle p_k:k\in\Bbb N\rangle$ be one of those Cauchy sequences. The whole idea of this construction of the real numbers is that $x$ should be the limit of the sequence $\sigma$. There’s a technical problem, however: the rational numbers $p_k$ aren’t actually in $\Bbb R$ when $\Bbb R$ is viewed as a set of equivalence classes of Cauchy sequences of rationals. However, if you’ve been shown this construction, you should have been shown that this $\Bbb R$ contains a copy of the rationals. 
Specifically, for each $k\in\Bbb N$ let $\pi_k$ be the constant sequence $\langle p_k,p_k,p_k,\ldots\rangle$; then $\bar\pi_k$ is the real number that corresponds to the rational $p_k$, where I write $\bar\tau$ for the equivalence class of a sequence $\tau$ of rationals. Thus, we expect that $\langle\bar\pi_k:k\in\Bbb N\rangle$ converges to $x=\bar\sigma$ in $\Bbb R$. To show this, we must show that for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $\bar d(\bar\pi_k,\bar\sigma)<\epsilon$ for each $k\ge m_\epsilon$, where $\bar d$ is the metric in $\Bbb R$. If you’ve done the rigorous development, you know that 
$$\bar d(\bar\pi_k,\bar\sigma)=\lim_{n\to\infty}|p_k-p_n|\;.$$
Since $\sigma$ is a Cauchy sequence, there is an $m_\epsilon\in\Bbb N$ such that $|p_\ell-p_n|<\frac{\epsilon}2$ whenever $\ell,n\ge m_\epsilon$. 


*

*Now show that $\bar d(\bar\pi_k,\bar\sigma)\le\frac{\epsilon}2<\epsilon$ whenever $k\ge m_\epsilon$ and conclude that $\langle\bar\pi_k:k\in\Bbb N\rangle$ is a sequence of rationals in $\Bbb R$ that converges to $x=\bar\sigma$.

A: Consider an infinite decimal expansion of $x$, and truncate it at rank $n$.  This gives a sequence $(a_n)$ that tends to $x$.
A: Let $x\in Q$ Then define the sequence of rationals by $x_n=x-\frac{1}{n}$.(Note that $x_n\in R$) Then $x_n\to x$ as $x\to \infty.$
Let $x\in Q^c$.Let $A_n=(x-\frac{1}{n},x+\frac{1}{n})$.There exists $x_n\in Q\cap A_n$ using the claim 1 below. So we have $x_n\to x$ as $n\to \infty$.
Claim 1:Let $p\ne q$ and $p,q\in R$. I will prove that $\exists x\in Q$ with $p<x<q$.
By Archimedean principle $\exists m\in R$ such that $m(q-p)>1\Rightarrow mq-mp>1$
Now as $mq-mp>1$ so $\exists n\in N$ such that $mp<n<mq\Rightarrow p<\frac{n}{m}<q$ and as $\frac{n}{m}\in Q$ so $x=\frac{n}{m}$ satisfies the claim.
Proof of Archimedean property,
Let $a>0$ and $b\in R$ and $S=\{n|n\in N\text{ and }na\le b\}$ We know that as $a>0$ so $na\to \infty $ as $n\to \infty$ so $|S|<\infty$ i.e. $S$ is finite.And as $S$ is finite so $\sup S$ exists. Now by choosing $m\in N, m>\sup S$ we have a natural no. such that $ma>b$ . This proves the archimedean property.
A: The real numbers are, or can be, defined using the type limit of limit you mention so there is nothing to prove, except that axiom A implies A.  
