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 Post subject: On a Cauchy sequencePosted: Fri Jul 07, 2017 6:32 am

Joined: Sat Nov 07, 2015 6:12 pm
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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.

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 Post subject: Re: On a Cauchy sequencePosted: Fri Jul 07, 2017 2:26 pm
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
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 $\displaystyle{a_n\stackrel{d}{\to}a}$.

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