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Let $(X,d)$ be a metric space. If $a_n$ is a sequence in $X$ and $a\in X$ such that $\frac{d(a_n,a)}{1+d(a_n,a)} \to 0$, then $a_n\to a$.

I am trying to prove this statement as review for an exam. I am not sure how to do it. I know that given $\epsilon >0$, since $\frac{d(a_n,a)}{1+d(a_n,a)} \to 0$, I can choose $N\in \mathbb{N}$ such that if $n\geq \mathbb{N}$, then $|\frac{d(a_n,a)}{1+d(a_n,a)} - 0| < \epsilon.$ I know that I need to show that there exists $N\in \mathbb N$ such that if $n\geq \mathbb N$, then $d(a_n,a) < \epsilon.$ Any suggestions?

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    $\frac{d(a_n,a)}{1+d(a_n,a)} < \epsilon$ implies $d(a_n,a) < \frac{\epsilon}{1-\epsilon}$. – mathworker21 Sep 30 '18 at 06:46
  • I did see this but I wasn't sure how to use it. Would it suffice to say that if $n\geq N$, then $d(a_n,a)<\frac{\epsilon}{1-\epsilon}<\epsilon$. So, $a_n\to a$. –  Sep 30 '18 at 06:50
  • it's not true that $\frac{\epsilon}{1-\epsilon} < \epsilon$. Here's the proof. Take $\epsilon > 0$. Take $\alpha > 0$ so that $\frac{\alpha}{1-\alpha} = \epsilon$ (i.e. $\alpha = \frac{\epsilon}{1+\epsilon}$). Then take $n \ge N$ so that $\frac{d(a_n,a)}{1+d(a_n,a)} < \alpha$. Then we get $d(a_n,a) < \frac{\alpha}{1-\alpha} = \epsilon$ for $n \ge N$. – mathworker21 Sep 30 '18 at 07:14

3 Answers3

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With $\epsilon<\dfrac12$ $$\frac{d(a_n,a)}{1+d(a_n,a)} =|\frac{d(a_n,a)}{1+d(a_n,a)} - 0| < \epsilon$$ then $$d(a_n,a)<\dfrac{\epsilon}{1-\epsilon}<\dfrac{\epsilon}{2}$$

Nosrati
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I would go a more general approach to show what is behind this convergence property. In fact, one can show that if $(X,d)$ is a metric space that then $\frac{d}{1+d}$ is an equivalent metric, in fact a bounded one. As equivalent metrics induce the same topology one can conclude the desired convergence.

Jonas Lenz
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Put $x_n = d(a_n,a)\implies 1-\dfrac{1}{1+x_n}\to 0\implies 1+x_n \to 1\implies x_n \to 0\implies a_n \to a$ as $n \to \infty$.

DeepSea
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