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Let $(x_n)$ be a sequence and $(y_n)$ be a Cauchy sequence. Suppose for all positive integers $n$, $$|x_n-y_n|< \frac{1}{n}$$ Prove that $(x_n)$ is also Cauchy.

My attempt: Suppose that $n \geq m$. Notice that $|x_n-x_m| \leq |x_n-y_n| + |y_n -y_m|+ |y_m-x_m|$

Since $(y_n)$ is a Cauchy sequence, for all $\epsilon >0$ , there exists $N \in \mathbb{N}$ such that for all $n \geq m \geq N$, $|y_n-y_m|< \epsilon$

Hence, we have $|x_n-x_m| \leq |x_n-y_n| + |y_n -y_m|+ |y_m-x_m| < \frac{1}{n} + \frac{1}{m}+ \epsilon =\frac{m+n}{nm}+ \epsilon < m+n+ \epsilon < \epsilon$ since $m,n \in \mathbb{N}$

Hence, $(x_n)$ is a Cauchy.

Is my proof valid?

EDIT: Choose $N$ such that $N=\frac{1}{\epsilon}$. Hence, for all $n \geq m \geq \frac{1}{\epsilon}$, we have $|x_n-x_m| < \frac{1}{n} + \frac{1}{m}+ \epsilon< 3 \epsilon$

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    $\begingroup$ How do you follow $m+n+\epsilon < \epsilon$? You should choose $n,m$ big enough for $\lvert y_n - y_m \rvert < \frac \epsilon 3$ and $\frac 1 n , \frac 1 m < \frac \epsilon 3$ $\endgroup$ – Stefan Apr 10 '13 at 19:19
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Not quite, you have $n+m + \epsilon < \epsilon$, which cannot hold if $m,n \ge 0$.

Choose $N$ large enough so that $\frac{1}{N} < \epsilon$. Then, if $m,n \ge N$, you have $|x_n-x_m| \leq |x_n-y_n| + |y_n -y_m|+ |y_m-x_m| < \frac{1}{n} + \frac{1}{m}+ \epsilon < 3 \epsilon$.

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  • $\begingroup$ ooh, thx. So I can choose $N$ as large as possible ? $\endgroup$ – Idonknow Apr 10 '13 at 19:24
  • $\begingroup$ No, you just need $N$ large enough so that if $m,n \ge N$, then $\frac{1}{m} < \epsilon$ and $\frac{1}{n} < \epsilon$. $\endgroup$ – copper.hat Apr 10 '13 at 19:30
  • $\begingroup$ i see. In this case, we have both $m$ and $n$ are larger than same thing, that is $1/ \epsilon$. If $m$ and $n$ are larger than two different terms, then which one should we choose ? Should we choose the bigger one ? $\endgroup$ – Idonknow Apr 10 '13 at 19:32
  • $\begingroup$ I don't understand your question. To prove Cauchy, you need to show that for all $\epsilon >0$, you can find $N$ such that if $m,n \ge N$, then the Cauchy condition holds. $\endgroup$ – copper.hat Apr 10 '13 at 19:37
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Hint. Apply the Archimedean Principle for $\frac{1}{n}$ and $\frac{1}{m}$.

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As $(y_n)$ is cauchy so it converges(let $\lim_{n\to \infty}y_n=y$) and the above inequality implies that $(x_n)$ also converges (that also to $y$). Now as any convergent sequence is cauchy so $(x_n)$ must also be cauchy.

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