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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 ?

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I am not sure I understand your strategy, or perhaps there is a typo. Suppose $a_j$ is just an infinite sequence of $5$s. Then the $b_j$ are all negative, and certainly can't converge to $5$. Or is another sequence supposed to be the new sequence? –  Potato Jun 20 '13 at 7:37
    
(I could post an answer if you like, but I think it is more instructive to let you work through your own ideas.) –  Potato Jun 20 '13 at 7:38
    
Presumably you are working in the reals, which are complete. So, can you find a sequence converging to $x$ that satisfies $x_n < x$? What number is $\frac{1}{n}$ away from $x$, and less than $x$? –  copper.hat Jun 20 '13 at 7:57
    
My idea was, I understand that ${1 \over j}$ is a sequence that converges to $0$ AND it's terms are always greater than 0. So I want to shift where the sequence converges to. However, if I use $x - \frac{1}{j}$, then $x$ may be irrational, which is not allowed in my case because the sequence must be rational. So I thought, if $\{x_j\} \to x$, then $\{x_j-\frac{1}{j}\} \to x$. The new problem is that now this new sequence is not always less than $x$. So I took the absolute value and negated the sequence so that each term is at most $x$ ... –  hyg17 Jun 20 '13 at 16:54
    
Thus, all I had to do is to convert all of the terms in $\{x_j\}$ which equaled zero into -1, so that now ALL terms are less than $x$. I should have mentioned, "if $x \ne 0$," there are only finite terms where $x_j = 0$ so that $\{x_j-\frac{1}{j}\} \to x$. –  hyg17 Jun 20 '13 at 16:57
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2 Answers 2

(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 $\lim_{m\to\infty} x_m=x$.

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Haha, thank you. Although I appreciate your help I don't know how this helps me because I am already aware of this. –  hyg17 Jun 20 '13 at 16:58
    
@hyg17: See my edit. –  Christian Blatter Jun 20 '13 at 18:46
    
Ah, so my approach is fine. That's good to hear. But I need all $x_n$ to be rational and less than $x$. I'm having trouble algebraically only proving that part. –  hyg17 Jun 20 '13 at 19:10
    
@hyg17: See my edit. –  Christian Blatter Jun 20 '13 at 19:26
    
I am so sorry to be anal, but I still haven't learned neither the decimal expansion nor the Dedekind cut for real numbers, so all I pretty much know is that real numbers are equivalent classes of a sequence of rationals. So I am not sure what you are saying. –  hyg17 Jun 20 '13 at 19:30
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Let $y_i = x-\max(|x-x_i|, \frac{1}{i})$.

Then $y_i < x$ and, since $|x-x_i| \to 0$, $y_i \to x$.

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Haha, I like the way everyone uses the same argument. I actually appreciate this idea and I understand it very well, however, I need the Cauchy sequence to be rational and that's where I am having most trouble. –  hyg17 Jun 20 '13 at 19:42
    
Once you have a $y_i < x$, you can find a rational between $y_i$ and $x$ as has been shown here before. –  marty cohen Jun 21 '13 at 5:33
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