Suppose $(x^{(n)})$ is a sequence in $\mathbb R^k$, $k \in \mathbb N$. From what I understand, $(x^{(n)})$ is a sequence of sequences $(x_1^{(n)}, x_2^{(n)}, x_3^{(n)}, ..., x_k^{(n)})$. Or in other words, $(x_j^{(n)})$ are terms of the sequence $(x^{(n)})$ for $j \in \mathbb N, 1\le j \le k$.
By the definition of a Cauchy sequence in $\mathbb R^k$, if $\epsilon \gt 0$, there exists $N$, such that $$m,n \gt N \implies d(x^{(m)}, x^{(n)}) \lt \epsilon$$ If $(x_j^{(n)})$ are terms of the sequence $(x^{(n)})$, then what is $(x^{(m)})$? By the definition of a cauchy sequence in $\mathbb R$, which is that if $\epsilon \gt 0$, there exists $N$, such that $$m,n > N \implies |s_n-s_m| < \epsilon$$ In this case, $s_m$ is a term in the sequence $s_n$, so that should mean that $(x^{(m)})$ too is a term of $(x^{(n)})$. But aren't $(x_1^{(n)}, x_2^{(n)}, x_3^{(n)}, ..., x_k^{(n)})$ the terms of $(x^{(n)})$? I realise that this question is stupid, but I doubt I'm gonna be able to study further on this topic unless I clear this up. Thank you for your time.
Edit: I am defining $d(x,y)$ as follows- $$d(x,y) = \left[\sum_{j=1}^k(x_j-y_j)^2\right]^{\frac{1}{2}} \text{ for } x,y \in \mathbb R^k$$
A sequence in $\mathbb{R}^k$ is not a sequence of sequences, it is a sequence of vectors. Usually, when we speak of sequences, it is implied that they have an infinite number of terms. In your definition of a Cauchy sequence, both $x^{(n)}$ and $x^{(m)}$ are vectors in $\mathbb{R}^k$, and $d(x^{(n)}, x^{(m)})$ is a real number, given, in the case of the Euclidean instance, by $$ d(x^{(n)},x^{(m)}) = \left(\sum_{j=1}^k(x^{(n)}_j-x^{(m)}_j)^2 \right)^{1/2}. $$
Since in finite dimensional spaces all norms are equivalent, you can also use distances like
$$ d(x^{(n)},x^{(m)}) = \sum_{j=1}^k|x^{(n)}_j-x^{(m)}_j| $$
or
$$ d(x^{(n)},x^{(m)}) = \max_{j=1, \cdots, k}|x^{(n)}_j-x^{(m)}_j|. $$