Euclidean Space is Banach Space

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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 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:

\(\displaystyle \size {x_n^{\paren k} - x_m^{\paren k} }\) \(=\) \(\displaystyle \paren {\paren {x_n^{\paren k} - x_m^{\paren k} }^2}^{\frac 1 2}\)
\(\displaystyle \) \(\le\) \(\displaystyle \paren {\sum_{k \mathop = 0}^m \paren {x_n^{\paren k} - x_m^{\paren k} }^2}^{\frac 1 2}\)
\(\displaystyle \) \(=\) \(\displaystyle \norm { {\bf x}_n - {\bf x}_m}_2\)

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

Then:

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

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

We have that for all $n > N$:

\(\displaystyle \norm {\mathbf x_n - \mathbf L}\) \(=\) \(\displaystyle \paren {\sum_{k \mathop = 1}^m \size {x^{\paren k}_n - x^{\paren k} }^2}^{\frac 1 2}\)
\(\displaystyle \) \(\le\) \(\displaystyle \paren {\sum_{k \mathop = 1}^m \frac {\epsilon^2} m}^{\frac 1 2}\)
\(\displaystyle \) \(=\) \(\displaystyle \epsilon\)

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

$\blacksquare$