A specific proof of inequality needed for an analysis problem. I am working on proving that there exists a Cauchy sequence approaching to $x$ where all the terms of $\{x_j\}$ is strictly less than $x$.
My plan is to construct a new sequence as follows.

Take any Cauchy sequence $\{a_j\} \to x$
Define $\{b_j\}$ equal to $\{-|a_j-\frac{1}{j}|\}$ where all $b_j \le x$.
Define $\{c_j\}$ such that if any term of $b_j = 0$ it is substituted by -1, that way all terms are now less than $x$.

I know how to show that the last part is valid, but I am stuck with a rather algebraic part where I need to show that if $\{a_j\}$ is Cauchy, $\{b_j\}$ also is.
Can someone help me out ?
 A: (i) A hint: Any sequence $(x_n)_{n\geq1}$ converging to $x$ is automatically a Cauchy sequence. It follows that
$$x_n:=x-{1\over n}\qquad(n\geq1)$$ does the job.
(ii) If you need rational $x_n$ the following might help: Assume $x>0$. Then $x$ possesses a decimal expansion $$x= a_0.a_1a_2a_3a_4\ldots$$ with $a_0\in{\mathbb N}$, $\>a_k\in \{0,\ldots,9\}$ $\>(k\geq1)$, and not ending with zeros. Now put
$$x_n:=a_0.a_1a_2a_3a_4\ldots a_n\qquad(n\geq1)\ .$$
(iii) If your real number $x$ is given (or represented) by a Cauchy sequence $(r_k)_{k\geq1}$ of rational numbers define a new sequence $(x_m)_{m\geq1}$ as follows: For each $m\geq1$ there is an $n=n(m)$ such that
$$|r_i-r_k|\leq {1\over m}\qquad(i,k\geq n(m))\ .$$
Put
$$x_m:=r_{n(m)}-{2\over m}\in {\mathbb Q}\qquad(m\geq1)\ .\tag{1}$$
Then
$$r_k\geq r_{n(m)}-{1\over m}=x_m+{1\over m}\qquad(k\geq n(m))$$
and therefore $$x=\lim_{k\to\infty} r_k\geq x_m+{1\over m}\ \qquad(m\geq1)\ .$$
It follows that $x_m<x$ for all $m\geq1$. Furthermore letting $m\to\infty$ in $(1)$ immediately shows that $\lim_{m\to\infty} x_m=x$.
A: Let $y_i = x-\max(|x-x_i|, \frac{1}{i})$.
Then $y_i < x$ and,
since $|x-x_i| \to 0$,
$y_i \to x$.
