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So i have a metric defined as: $$d_{f}(x,y)=|f(x)-f(y)|$$ Where $f: \mathbb{R}\to\mathbb{R}$ is defined as: $$f(x)=\begin{cases}x-1&\mbox{if }x< 0\\0&\mbox{if }x= 0.\\x+1&\mbox{if }x> 0.\end{cases}$$

So i have three sequences: $$a_{n}=\frac{(-1)^n}{n}, b_{n}=\frac{1}{n}, c_{n}=2+\frac{(-1)^n}{n}$$

Andi need to check if they are either convergent, either Cauchy sequences, or perhaps none. Also this will help getting an answer to the question about completeness of this metric space $(\mathbb{R},d_{f})$

So i know that the way to investigate convergence and Cauchy sequences is simmilar to, when using normal absolute value as metric, but i really need help with this special metric, because it's giving me problems how to even start. Any help would be appreciated.

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I will take $\{b_n\}$ as an example. The distance between $b_m$ and $b_n$ is, by definition,

$$d_f (b_m,b_n) = |f(b_m)-f(b_n)|$$

I am assuming that $n\in \mathbb{N}$. If that is the case, then notice that for $m,n$ sufficiently large, we really have

$$|f(b_m) - f(b_n)|=|b_m+1-b_n-1|=|b_m - b_n|$$

Of course, this is just

$$|b_m - b_n| = |\frac{1}{m} - \frac{1}{n}|$$

Can you take it from here?

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  • $\begingroup$ Yeah i can take it from here, but also...i think that $a_n$ and/or $b_n$ will be the reasons for not bein a complete metric space , i just think it's like that $\endgroup$ – MathIsTheWayOfLife Sep 3 '16 at 11:21
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You may check by hand. So e.g. show that $(b_n)$ is Cauchy, but that there is no limit, or that $(a_n)$ is not even Cauchy (why?).

Alternatively you may notice that the map $f$ itself provides an isometric bijection between $({\Bbb R},d_f)$ and $$ A=(-\infty,-1) \cup \{0\} \cup (1,+\infty)$$ equipped with the usual topology from ${\Bbb R}$. And isometric bijections preserve the relevant properties of sequences (being Cauchy, convergent), thus in particular completeness.

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