# Proving a sequence is bounded away from zero

I was trying to go through the book Analysis I.

The chapter I am on is trying to build up the real numbers as Cauchy sequences of the rationals, and prove some properties. One of these properties is the trichotomy of reals - they are positive, negative or zero.

This is asked as a problem. There is a hint associated with the question.

If $x$ is not zero, then $x$ is the limit of some sequence $(a_n)_{n=1}^{\infty}$, which is not equivalent to the sequence $(0)_{n=1}^{\infty}$. Use this to prove that the sequence is either positively or negatively bounded away from zero.

Here is my take on it. We know the sequence is not equivalent to zero. This means for some $M$ and some $\epsilon_1$,

$$|a_M - 0| > \epsilon_1$$ $$|a_M| > \epsilon_1$$

Also because $x$ is real, the sequence $(a_n)_{n=1}^{\infty}$ is Cauchy, which means $$|a_m - a_n| < \epsilon \;\;\;\;\forall n,m \geq N, \epsilon > 0$$

Using this, we need to show that

$$|a_n| > \delta \;\;\;\;\;\; \delta > 0, n \geq K$$

for some $\delta$ and some $K$.

The idea is to find some $M$ and $\epsilon_1$ such that $M > N$ and $\epsilon < \epsilon_1$, because then we can show

$$|a_m - a_n| < \epsilon \;\;\;\;\forall n,m \geq N$$ Specifically, $$|a_M - a_n| < \epsilon \;\;\;\;\forall n \geq N$$ $$|a_M| - |a_n| < \epsilon \;\;\;\;\forall n \geq N$$

Now, $|a_M| > \epsilon_1$ or $\epsilon_1 < |a_M|$

$$\epsilon_1 - |a_n| < \epsilon \;\;\;\;\forall n \geq N$$

$$|a_n| > \epsilon_1 - \epsilon \;\;\;\;\forall n \geq N$$

This means that the sequence is eventually bounded away from zero. And is therefore eventually positive or negative.

What this proof rests on is the fact that we can find some $M$ and $\epsilon_1$ such that $M > N$ and $\epsilon < \epsilon_1$. I just don't know how to prove that. Could someone help?

• @Kavi the sequence you are giving in not cauchy. – Shervin Sorouri Jun 12 '18 at 9:41
• You have already shown that the sequence is bounded away from zero as far absolute value is concerned. Now show that the sequence maintains a constant sign (same as sign of $x$). Can you do that by taking by $\epsilon=|x|/2$? – Paramanand Singh Jun 12 '18 at 10:03
• @ParamanandSingh The thing is I cannot use $x$(or $x/2$) as the epsilon, as we are dealing with sequences of rationals bounded by rationals. But I get the intuition of what you are saying - use some $\epsilon$ to show that all elements in the sequence have the same sign eventually. I just can't pin it down rigorously. As I say in the last line $\\$ >this proof rests on is the fact that we can find some $M$ and $\epsilon_1$ such that $M > N$ and $\epsilon < \epsilon_1$. I just don't know how to prove that – wamiq reyaz Jun 12 '18 at 12:28
• Then take $\epsilon=\epsilon_1/2$ and show that $a_n$ has same sign as that of $a_M$. – Paramanand Singh Jun 13 '18 at 2:00

The sequence $\{a_n\}$ is not equivalent to $\{0\}$ which implies that there is a rational number $\epsilon_1>0$ such that there are infinitely many positive integers $M$ with $|a_M-0|>\epsilon_1$ ie $|a_M|>\epsilon_1$.
Now take $\epsilon=\epsilon_1/2$ and since the sequence $\{a_n\}$ is Cauchy it follows that there is a positive integer $N$ such that $|a_n-a_m|<\epsilon$ whenever $n\geq N\leq m$. By the last paragraph we can choose an $M>N$ and then set $m=M$ to get $$|a_n-a_M|<\frac {\epsilon_1}{2}$$ for all $n\geq N$.
Using the above inequality and $|a_M|>\epsilon_1$ you should be able to prove that $|a_n|>\epsilon_1/2$ and $a_n$ has same sign as that $a_M$. Take the cases $a_M<0$ and $a_M>0$ separately.
• Thanks. The important thing I realized later is that no matter which part of the sequence I look at I can always find an $\epsilon_1$ and a corresponding $M$. The rest can be easily done. Thanks again. – wamiq reyaz Jun 13 '18 at 8:35
The definition of Cauchy sequence provides you with a M for every $\epsilon >0$. Therefore you can choose an $\epsilon < \epsilon_1$. For this $\epsilon$ there exists an $M_1$. For your $M$, take the larger of $N$ and $M_1$.