Euclidean Space is Banach Space

Theorem
Let $m$ be a positive integer.

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

Corollary
The complex plane, along with the metric induced by the norm given by the complex modulus, forms a Banach space over $\C$.

Proof
The Euclidean space $\R^m$ is a vector space over $\R$.

That the norm axioms are satisfied is proven in Euclidean Metric is a Metric.

Then we have that a Euclidean space is a complete metric space.

The result follows by the definition of a Banach space.

Proof of Corollary
The complex plane $\C$ is a vector space over $\C$.

That the norm axioms are satisfied is proven in Modulus of Product and Triangle Inequality.

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.

By the theorem, $\C$ is a complete metric space.

Hence the result.