Euclidean Space is Banach Space/Proof 2

Theorem
Let $m$ be a positive integer.

Then the Euclidean space $\R^m$, along with the Euclidean norm, forms a Banach space over $\R$.

Proof
By definition, Euclidean norm is the same as $p$-norm with $p = 2$.

Let $\sequence {\mathbf x_n}_{n \mathop \in \N} = \sequence {\tuple {x_n^{\paren 1}, x_n^{\paren 2}, \ldots, x_n^{\paren m}}}_{n \mathop \in \N} $ be a Cauchy sequence in $\R^m$.

Let $k \in \N_{> 0} : k \le m$.

Then:

Hence, $\sequence {x_n^{\paren k} }_{n \mathop \in \N}$ is a Cauchy sequence in $\R$.

Then:


 * $\displaystyle \lim_{n \mathop \to \infty} x_n^{\paren k} = L^{\paren k}$

Let $\mathbf L = \tuple {L^{\paren 1}, \ldots, L^{\paren m}} \in \R^m$

We have that for all $n > N$:

Therefore, $\sequence {\mathbf x_n}_{n \mathop \in \N}$ converges to $\mathbf L$ in $\struct {\R^m, \norm {\, \cdot \,}_2}$.