Euclidean Space is Complete Metric Space

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Theorem

Let $m$ be a positive integer.

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

$\blacksquare$


Sources