How to prove that the following statements are equivalent: $\lim_{n\to \infty} a_n = a$ and $\lim_{x\to 0} \bar{\alpha}(x) = a$

Question

In an assignment for the course Real-Analysis I need to proof that that the following statements are equivalent:

1. $\lim_{n\to \infty} a_n = a$
2. $\lim_{x\to 0} \bar{\alpha}(x) = a$

Given that $(W, d_w)$ is a metric space and $\alpha: \mathbb{N} \longrightarrow W$ and $\alpha(n) = a_n$ a row in $W$. We can use the function $\bar{\alpha}:\mathbb{R} \supset \longrightarrow W$ with:

$$Dom(\bar{\alpha}) = \{\frac{1}{n+1} : n \in \mathbb{N}\} \subset \mathbb{R}$$

and

$$\bar{\alpha}(x) = \alpha (\frac{1}{x} -1), x \in Dom(\bar{\alpha})$$

Let $a \in W$ and prove that the following statements are equivalent:

1. $\lim_{n\to \infty} a_n = a$
2. $\lim_{x\to 0} \bar{\alpha}(x) = a$

However, me and my fellow students have been trying to figure it out for some time now, but we can't...

Answer 1

Let's write $x=\frac{1}{n+1}~~~n \in \mathbb{N}$. $\bar{\alpha}(\frac{1}{n+1})=\alpha(n)=a_n~~~\forall n$.

$1. \Rightarrow 2.$ For given $\epsilon >0~~~\exists N \in \mathbb{N}$ such that $d_w(a_n,a)<\epsilon$ for all $n >N$. Then for $\frac{1}{n+1}< \frac{1}{N+1}$ ($|x|<\delta$),

$d_w(\bar\alpha(\frac{1}{n+1}),a)\leq d_w(\bar\alpha(\frac{1}{n+1}),a_n)+d_w(a_n,a)<\epsilon$

$2. \Rightarrow 1.$ Given $\epsilon >0~~~\exists\delta>0$ such that $d_w(\bar\alpha(\frac{1}{n+1}),a)<\epsilon$ whenever $\frac{1}{n+1}<\frac{1}{M+1}<\delta$ (for some large $M \in \mathbb{N}$). Then for $n >M$, $d_w(a_n,a)\leq d_w(\bar\alpha(\frac{1}{n+1}),a_n)+d_w(\bar\alpha(\frac{1}{n+1}),a)<\epsilon$