I need some help with this problem, I've seen that there is a similar problem about proving that Euclidean Space is a complete metric space, but I haven't learn about Euclidean or metric spaces and its notation, so the answer in that question isn't helping me.
A sequence $(a_k)$ in $\mathbb{R}^n$ is a Cauchy sequence if to each $\varepsilon>0$ there corresponds $K\in\mathbb{N}$ such that $$\Vert a_k-a_l\Vert<\varepsilon$$ whenever $k,l>K$. Prove that a Cauchy sequence in $\mathbb{R}^n$ is convergent. Hint apply the ideas in the proof of Theorem 2.2.7 and 1.6.4
Ok, so I tried to do it like this:
From the theorem 2.2.7, we know that if $(a_k)$ is a sequence in $\mathbb{R}^n$ it converges $a_k\rightarrow a \iff a_{ki}\rightarrow a_i, \ \forall i=1,...,n.$ And $(a_l)$ is a sequence in $\mathbb{R}^n$ it converges $a_l\rightarrow a\iff a_{li}\rightarrow a_i, \ \forall i=1,...,n \ $, where $a_k=(a_{k1},...,a_{kn})\forall k\in\mathbb{N}$, $a=(a_1,...,a_n)\in\mathbb{R}^n$ and $a_l=(a_{l1},...,a_{ln})\forall l\in\mathbb{N}$, $
We know that if the sequence $(a_k)$ converges, then $\forall\varepsilon>0,\ \exists \ K\in\mathbb{N}$ such that $k>K$ $$\Vert a_k-a\Vert<\frac{\varepsilon}{2}$$ and $\exists l\in\mathbb{N}$ such that $l>K$ $$\Vert a_l-a\Vert<\frac{\varepsilon}{2}$$ $\Rightarrow \Vert a_k-a_l\Vert=\Vert a_k-a_l-a+a\Vert \leq\Vert a_k-a\Vert + \Vert a-a_l\Vert < \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$
Thus $\Vert a_k-a_l\Vert<\varepsilon$.
Is my proof correct or not? In case not, how can I prove this correctly? Thanks.