# Hilbert Sequence Space is Complete Metric Space

## Theorem

Let $A$ be the set of all real sequences $\left\langle{x_i}\right\rangle$ such that the series $\displaystyle \sum_{i \mathop \ge 0} x_i^2$ is convergent.

Let $\ell^2 = \left({A, d_2}\right)$ be the Hilbert sequence space on $\R$.

Then $\ell^2$ is a complete metric space.

## Proof

We have that Hilbert Sequence Space is Metric Space.

It remains to be shown that it is complete.

Recall that from Real Number Line is Complete Metric Space, $\R$ is a complete metric space.

Let $x^1, x^2, x^3, \ldots$ be a Cauchy sequence $\ell^2$.

Then for each $i \in \N_{>0}$, we have that $\left\langle{ {x_i}^j}\right\rangle_{j \mathop \in \N_{>0} }$ is a Cauchy sequence in the complete metric space $\R$.

Thus $\left\langle{ {x_i}^j}\right\rangle_{j \mathop \in \N_{>0} }$ converges to a point in $\R$, say $x_i$.

Then if $x = \left\langle{x_i}\right\rangle$, the points $x - x_j$ eventually belong to $H$.

So $x = \left({x - x_j}\right) + x^j$ must also belong to $H$.

Thus $d \left({x, x^j}\right) = 0$.

Hence the result.

$\blacksquare$