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On a Cauchy sequence

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Tolaso J Kos
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On a Cauchy sequence


Post by Tolaso J Kos » 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.
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Papapetros Vaggelis
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Re: On a Cauchy sequence


Post by Papapetros Vaggelis » 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


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

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


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