#2

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

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

**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 \(\displaystyle{a_n\stackrel{d}{\to}a}\).