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### On a Cauchy sequence

Posted: Fri Jul 07, 2017 6:32 am
Let $\mathbb{R}^+ =\{ x \in \mathbb{R}: x>0\}$. Endow it with the metric

$${\rm d}(x, y) = \left| \frac{1}{x} - \frac{1}{y} \right|$$
1. Show that the sequence $a_n=n$ is a Cauchy one.
2. Is the sequence $\frac{1}{n}$ a Cauchy one?
3. Show that any sequence $a_n$ in $\mathbb{R}^+$ converges in $\mathbb{R}^+$ in the metric ${\rm d}$ above if and only if it converges in $\mathbb{R}$ in the standard metric $|x-y|$ and that the limits in the two cases are equal.

### Re: On a Cauchy sequence

Posted: Fri Jul 07, 2017 2:26 pm
It's obvious that $$\displaystyle{d}$$ is a metric.

i. Let $$\displaystyle{\epsilon>0}$$. There exists $$\displaystyle{n_0\in\mathbb{N}}$$ such

that $$\displaystyle{\dfrac{2}{n_0}<\epsilon}$$. Then, for every $$\displaystyle{n\,,m\in\mathbb{N}}$$

such that $$\displaystyle{n\,,m\geq n_0}$$ holds

$$\displaystyle{d(a_n,a_m)=\left|\dfrac{1}{n}-\dfrac{1}{m}\right|\leq \dfrac{1}{n}+\dfrac{1}{m}\leq \dfrac{1}{n_0}+\dfrac{1}{n_0}=\dfrac{2}{n_0}<\epsilon}$$

So, the sequence $$\displaystyle{\left(a_n=n\right)_{n\in\mathbb{N}}}$$ is a Cauchy one.

ii. The answer is negative. If $$\displaystyle{b_n=\dfrac{1}{n}\,,n\in\mathbb{N}}$$,

we observe that

$$\displaystyle{d(b_n,b_{n+1})=\left|n-(n+1)\right|=1\,,\forall\,n\in\mathbb{N}}$$.

iii. Let $$\displaystyle{\left(a_{n}\right)_{n\in\mathbb{N}}}$$ be a sequence of $$\displaystyle{\mathbb{R}^{+}}$$.

Suppose that $$\displaystyle{a_{n}\stackrel{d}{\to} a>0}$$, that is

$$\displaystyle{d(a_n,a)\to 0\iff \left|\dfrac{1}{a_n}-\dfrac{1}{a}\right|\to 0}$$, which means that

$$\displaystyle{\dfrac{1}{a_n}\stackrel{|\cdot|}{\to}\dfrac{1}{a}\implies a_n\stackrel{|\cdot|}{\to}a}$$

Similarly,

if $$\displaystyle{a_n\stackrel{|\cdot|}{\to}a}$$, then $$\displaystyle{a_n\stackrel{d}{\to}a}$$.