Euclidean Space is Complete Metric Space

Theorem
Let $m$ be a positive integer.

Then the Euclidean space $\R^m$, along with the Euclidean metric, forms a complete metric space.

Corollary
The complex plane, along with the metric induced by the norm given by the complex modulus, forms a complete metric space.

Proof
Let $\left \langle {\left( {x_{n,1}, x_{n,2}, \ldots, x_{n,m}} \right)} \right \rangle_{n \in \N} $ be a Cauchy sequence in $\R^m$.

Let $\epsilon > 0$.

Then $\dfrac \epsilon m > 0$.

We have that Real Number Line is Complete Metric Space.

Therefore:
 * $\exists y_1, y_2, \ldots, y_m \in \R$ and $N_1, N_2, \ldots, N_m \in \N$ (depending on $\epsilon$)

such that:
 * $\forall k \in \N: 1 \le k \le m: \forall n_k > N_k: \left\vert{x_{n,k} – y_k}\right\vert < \dfrac \epsilon m$

From Euclidean Space is Normed Space:


 * $\displaystyle \left \Vert{\left({x_{n,1}, x_{n,2}, \ldots, x_{n,m}}\right) - \left({y_1, y_2, \ldots, y_m}\right)}\right\Vert \le \sum_{k \mathop = 1}^m \left\vert{x_{n, k} – y_k}\right\vert < \epsilon$

Hence the Euclidean space is a complete metric space.

Proof of Corollary
Let $z = x + iy$ be a complex number, where $x, y \in \R$.

Now, we can identify the complex number $z$ with the ordered pair $\left( x, y \right) \in \R^2$.

The norm on $\C$ given by the complex modulus is then identical to the Euclidean norm on $\R^2$.

Therefore, metric on $\C$ induced by the norm given by the complex modulus is identical to the Euclidean metric on $\R^2$ induced by the Euclidean norm.

Hence $\C$ is a complete metric space.