# Limitations that make a simple euclidean space a Hilbert space

From what I understood is that a complex vector space $\mathbb{V}\cap\mathbb{C}^{\infty}$ becomes a Hilbert space if it is orthonormal, if the inner product of all vectors$(\left|x\right>,\left|y\right>)$ is defined and if the space is complete. From what I understood, completeness implies that the sum of vectors should converge towards a vector of finite length. Basically, what I understand is that $$||\sum_{k=0}^{\infty}\left|v_k\right>||<\infty$$ But it's not difficult for a mathematician to figure out that I could easily construct vectors in euclidean space that disobey this ruling, (but wikipedia says that one of the most famous examples of Hilbert space is the euclidean space). For example let met have a vector $0<\left|||A\right>||<\infty$ such that and let me define $\left|x_n\right>=\left(\frac{1}{n}\right)\left|A\right>$. This vector series wouldn't converge, then why is euclidean space in $\mathbb{R}^3$ a Hilbert space?

• A metric space is complete if every Cauchy sequence converge. Obviously we can have many non-convergent non-Cauchy sequences. Dec 6 '16 at 9:16

It seems that there is some confusion here.

First, note that the notion of ''orthonormal'' does not apply to a vector space but to a basis, i.e. a set of lineraly independent vectors that span the vector space.

More: the notion of orthogonality of vectors can be defined only if the space has an inner product. In this case the vector space is called a pre-Hilbert space, and we can prove that, the inner product define a norm, so the space is also a normed space (called a pre-Banach space) and from this norm we can define a distance so that the space is a metric space.

In a metric space we can define a notion of convergence for sequences of elements (vectors) and we can define Cauchy's sequences.

The (metric) space is called complete if all the Cauchy's sequences are convergent to a limit in the space.

( Note that the sequence $|x_n\rangle$ in OP is not a Cauchy sequence).

In this case, if the distance comes from a norm and this comes from an inner product, the space is also a normed Banach space and a Hilbert space.

Since there is a general way to complete any metric space, using the equivalence classes of Cauchy sequences, we can always complete a pre-Hilbert space, i.e. : any vector space with an inner product can be ''promoted'' to an Hilbert space with a standard procedure of completion.

For understanding what it means, you need to look at an example of a non-complete normed vector space :

Let $E,\|.\|_2$ bet the set of real (or complex) sequences ending with infinitely many zeros, with the $\|.\|_2$ norm $\|y\|_2 = \sqrt{\sum_n |y(n)|^2}$.

You can easily check it is a normed vector space.

Then let $x_k(n) = \begin{cases}\frac{1}{n^2} \text{ if } n \le k\\ 0 \text{ otherwise} \end{cases}$

since $\|x_k-x_{k'}\|_2 = \sum_{n\in (k,k']} \frac{1}{n^2}$ you see that $(x_k)_{k \in \mathbb{N}}$ is a Cauchy sequence in $E,\|.\|_2$, but $x_k$ converges in the $l^2$ norm to the sequence $y_n = \frac{1}{n^2}$ which is not in $E$.

The completion of $E,\|.\|_2$ is the Hilbert space $l^2$ containing all the sequence with finite $\|.\|_2$ norm.

$l^2$ is a Hilbert space because its norm comes from a inner product $$\|y\|_2^2 = \langle y,y \rangle \qquad \quad \langle x,y \rangle = \sum_n x_n y_n$$ (or $\langle x,y \rangle = \sum_n x_n \overline{y_n}$ if you are considering the complex sequences)