# Hilbert Sequence Space is 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 metric space.

## Proof 1

$\ell^2$ is a particular instance of the general $p$-sequence space $\ell^p$.

Hence $p$-Sequence Space of Real Sequences is Metric Space can be applied directly.

$\blacksquare$

## Proof 2

By definition of the Hilbert sequence space on $\R$:

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.

Then $\ell^2 := \left({A, d_2}\right)$ where $d_2: A \times A: \to \R$ is the real-valued function defined as:

$\displaystyle \forall x = \left\langle{x_i}\right\rangle, y = \left\langle{y_i}\right\rangle \in A: d_2 \left({x, y}\right) := \left({\sum_{k \mathop \ge 0} \left({x_k - y_k}\right)^2}\right)^{\frac 1 2}$

From Convergence of Square of Linear Combination of Sequences whose Squares Converge we have that $\displaystyle \sum_{k \mathop \ge 0} \left({x_k - y_k}\right)^2$ does actually converge.

### Proof of $M1$

 $\displaystyle d_2 \left({x, x}\right)$ $=$ $\displaystyle \left({\sum_{k \mathop \ge 0} \left({x_k - x_k}\right)^2}\right)^{\frac 1 2}$ Definition of $d_2$ $\displaystyle$ $=$ $\displaystyle \left({\sum_{k \mathop \ge 0} 0^2}\right)^{\frac 1 2}$ $\displaystyle$ $=$ $\displaystyle 0$

So axiom $M1$ holds for $d_2$.

$\Box$

### Proof of $M2$

Let $z = \left\langle{z_i}\right\rangle \in A$.

 $\displaystyle d_2 \left({x, y}\right) + d_2 \left({y, z}\right)$ $=$ $\displaystyle \left({\sum_{k \mathop \ge 0} \left({x_k - y_k}\right)^2}\right)^{\frac 1 2} + \left({\sum_{k \mathop \ge 0} \left({y_k - z_k}\right)^2}\right)^{\frac 1 2}$ Definition of $d_2$ $\displaystyle$ $\ge$ $\displaystyle \left({\sum_{k \mathop \ge 0} \left({x_k - z_k}\right)^2}\right)^{\frac 1 2}$ Minkowski's Inequality for Sums: index $2$ $\displaystyle$ $=$ $\displaystyle d_2 \left({x, z}\right)$ Definition of $d_2$

So axiom $M2$ holds for $d_2$.

$\Box$

### Proof of $M3$

 $\displaystyle d_2 \left({x, y}\right)$ $=$ $\displaystyle \left({\sum_{k \mathop \ge 0} \left({x_k - y_k}\right)^2}\right)^{\frac 1 2}$ Definition of $d_2$ $\displaystyle$ $=$ $\displaystyle \left({\sum_{k \mathop \ge 0} \left({y_k - x_k}\right)^2}\right)^{\frac 1 2}$ Definition of Absolute Value $\displaystyle$ $=$ $\displaystyle d_2 \left({y, x}\right)$ Definition of $d_2$

So axiom $M3$ holds for $d_2$.

$\Box$

### Proof of $M4$

 $\displaystyle x$ $\ne$ $\displaystyle y$ $\displaystyle \implies \ \$ $\, \displaystyle \exists k \in \N: \,$ $\displaystyle x_k$ $\ne$ $\displaystyle y_k$ $\displaystyle \implies \ \$ $\displaystyle \left({x_k - y_k}\right)^2$ $>$ $\displaystyle 0$ $\displaystyle \implies \ \$ $\displaystyle \left({\sum_{k \mathop \ge 0} \left({x_k - y_k}\right)^2}\right)^{\frac 1 2}$ $>$ $\displaystyle 0$ $\displaystyle \implies \ \$ $\displaystyle d_2 \left({x, y}\right)$ $>$ $\displaystyle 0$ Definition of $d_2$

So axiom $M4$ holds for $d_2$.

$\blacksquare$