Euclidean Space is Banach Space

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 1

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

That the norm axioms are satisfied is proven in Euclidean Space is Normed Vector Space.

Then we have Euclidean Space is Complete Metric Space.

The result follows by the definition of a Banach space.

$\blacksquare$

Proof 2

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:

\(\ds \size {x_n^{\paren k} - x_m^{\paren k} }\)

\(=\)

\(\ds \paren {\paren {x_n^{\paren k} - x_m^{\paren k} }^2}^{\frac 1 2}\)

\(\ds \)

\(\le\)

\(\ds \paren {\sum_{k \mathop = 0}^m \paren {x_n^{\paren k} - x_m^{\paren k} }^2}^{\frac 1 2}\)

\(\ds \)

\(=\)

\(\ds \norm { {\bf x}_n - {\bf x}_m}_2\)

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

Then:

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

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

We have that for all $n > N$:

\(\ds \norm {\mathbf x_n - \mathbf L}\)

\(=\)

\(\ds \paren {\sum_{k \mathop = 1}^m \size {x^{\paren k}_n - x^{\paren k} }^2}^{\frac 1 2}\)

\(\ds \)

\(\le\)

\(\ds \paren {\sum_{k \mathop = 1}^m \frac {\epsilon^2} m}^{\frac 1 2}\)

\(\ds \)

\(=\)

\(\ds \epsilon\)

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

$\blacksquare$