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Is the following statement true:

Let $(X,d)$ be a metric space and $(x_n)_n \in X ^\mathbb{N}$. Then:

$\exists m \in \mathbb{N}:d(x_m,x_n)\to 0$ when $n \to \infty $ in $(\mathbb{R}, d_E) \implies (x_n)_n$ is a Cauchy sequence

My attempt:

Let $\epsilon >0$ be arbitrary. Choose $n_0$ such that $\forall n > n_0: d(x_m, x_n) < \epsilon/2$.

Now, for $p,q > n_0$, we have: $d(x_p,x_q) \leq d(x_m, x_p) + d(x_m,x_q) < \epsilon/2 + \epsilon/2 = \epsilon$

Hence, $(x_n)_n$ is Cauchy.

Now, my actual question:

In a certain proof, it is stated that for $0 < n < m$, we have:

$$d(x_m,x_n) \leq \frac{k^n}{1-k}d(x_1,x)$$ where $0 <k < 1$

Can I use the claim I made to conclude that $(x_n)_n$ is a Cauchy sequence, because the right hand side goes to $0$?

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  • $\begingroup$ I don't understand your question. Do you want to know if your attempt is correct? What does the "actual question" has to do with the rest? $\endgroup$
    – Arnaud D.
    Nov 24, 2017 at 15:33
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    $\begingroup$ Not directly. In the claim, you have $\lim_{n\to\infty} d(x_m,x_n) = 0$ for a fixed $m$, in your inequality you have the constraint $n < m$, and the right hand side depends on $n$ but not on $m$. $\endgroup$
    – Daniel Fischer
    Nov 24, 2017 at 15:39
  • $\begingroup$ @DanielFischer How can I conclude that $(x_n)_n$ is a cauchy sequence then, given that for $0 < n < m$ we have $d(x_m,x_n) = \frac{k^n}{1-k}d(x_1,x)$? $\endgroup$
    – user370967
    Nov 25, 2017 at 8:46
  • $\begingroup$ Choose $m = n+1$ and consider $$\sum_{n = N}^{\infty} d(x_{n+1},x_n)\,.$$ $\endgroup$
    – Daniel Fischer
    Nov 25, 2017 at 10:18
  • $\begingroup$ In my course, we didn't handle series yet. $\endgroup$
    – user370967
    Nov 25, 2017 at 10:34

1 Answer 1

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The LHS is not negative, so yes.

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  • $\begingroup$ So everything I did was correct? $\endgroup$
    – user370967
    Nov 24, 2017 at 11:38
  • $\begingroup$ @ajotatxe Why is that? Suppose that the sequence is constant. Then the LHS holds (for every $m$ in fact). $\endgroup$
    – José Carlos Santos
    Nov 24, 2017 at 11:39
  • $\begingroup$ @JoséCarlosSantos ajotaxe's argument also applies to a constant sequence. $\endgroup$
    – Arnaud D.
    Nov 24, 2017 at 15:30

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