Given a metric space $(\Bbb N,d)$ where $d(x,y)= |\frac{1}{x}-\frac{1}{y}|.$

I need to prove that for a sequence $(n_j)_{j \in \Bbb N} \in \Bbb N$, (as $j \rightarrow \infty$, $n_j \rightarrow \infty$) $\Rightarrow$ this sequence is a Cauchy sequence.

Here is the definition of Cauchy seuence:

Let $(x_n)^∞ _{n=1} ⊂ X$ where $(X, d)$ is a metric space. Then $(x_n)$ is a Cauchy sequence if for every $\varepsilon > 0$ there exists an integer $N$ such that $m, n ≥ N ⇒ d(x_m, x_n) < \varepsilon$ .

By definition, I need an explicit $N$ for every $\varepsilon >0$ to prove that. I have spent some time on this question but did not make much progress. Can someone help me, please?

  • 2
    $\begingroup$ Well. Do it. If $n,m > N $ what does that say about $d (n,m)=|\frac 1n - \frac 1m|$? Remind yourself what the definition of $n_j\rightarrow \infty $ is. $\endgroup$ – fleablood Apr 1 '17 at 4:23
  • $\begingroup$ @fleablood I do know the fact that as $j\rightarrow \infty$ we have $d(n,m)\rightarrow 0$. But now I need an explicit construction for $N$, according to the definition of the Cauchy sequence. $\endgroup$ – PropositionX Apr 1 '17 at 4:31
  • $\begingroup$ If $e > 0$ what do m,n have to be so that $|1/m - 1/n| < e $ always? Hint, use $|1/m - 1/n| \le 1/m + 1/n $ or $|1/m -1/n| < min (1/m,1/n) $. $\endgroup$ – fleablood Apr 1 '17 at 4:41

Take $\epsilon > 0$. Take $M$ large so that $\frac{1}{M} < \epsilon$. Take $N$ large so that $\forall j \ge N, n_j \ge M$ which is possible since $n_j \to \infty$. Then, for $j,k \ge N$, $d(n_j,n_k) = |\frac{1}{n_j}-\frac{1}{n_k}| \le \max(\frac{1}{n_j},\frac{1}{n_k}) \le \frac{1}{M} < \epsilon$.

  • $\begingroup$ A typo maybe. But you have to edit your second sentence. $\endgroup$ – Juniven Apr 1 '17 at 4:44
  • $\begingroup$ Are you sure that $\frac{1}{M}>\epsilon$ in your second sentence? $\endgroup$ – Juniven Apr 1 '17 at 5:54
  • $\begingroup$ @mathworker21 In you second sentence, it should be $\frac{1}{M} < \varepsilon$. $\endgroup$ – PropositionX Apr 1 '17 at 6:36
  • $\begingroup$ Ah, I read the third statement instead of the second. Thank you very much! $\endgroup$ – mathworker21 Apr 1 '17 at 7:58

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