# Euclidean space without orthonormal basis

Problem Give an example of a nonseperable Euclidean space which has no orthonormal basis.

My Argument I know if a Euclidean space $R$ has at most countable basis, then it has an orthonormal system. Moreover i know every linear space has Hamel basis. So, i am looking for a Euclidean space with uncountable basis.

Trouble How can I construct uncountable basis? Its just too many elements to deal with. Could you please help me with this?

Edit By Euclidean space i mean, a linear space with scalar product. Not necessarily complete or seperable like Hilbert spaces.

• What do you mean exactly with Euclidean space? Personally, I use that terminology only to denote $\mathbb{R}^n$, $n<\infty$. Jan 13, 2014 at 11:51
• @DavidMitra Although those are called orthonormal basis, they are not bases in the usual sense of a vector space. You need infinite sums of them to get the whole space. They generate a space that is only dense in the whole space. The OP might be asking for existence of an actual basis that is also orthonormal.
– user119256
Jan 13, 2014 at 11:54
• @iamvegan Already taking a separable, infinite dimensional Hilbert space gives you the answer. Once you have constructed an orthogonal system $S$ that spans a set dense in the whole space, you cannot produce another vector that is orthonormal to all of them. Such a vector would be zero. On the other hand an infinite dimensional Hilbert space will have an uncountable Hamel basis. Therefore the countable orthonormal system $S$ is still not spanning (in the sense of linear algebra) the whole space.
– user119256
Jan 13, 2014 at 12:01
• Well, if that is true, then you don't need the word nonseperable. Separable Hilbert space does not have countable basis, and any orthonormal system is countable... Jan 13, 2014 at 12:02
• @iamvegan First notice that a Hilbert space basis is not the same as a basis (called Hamel basis to distinguish them) of the same space as a vector space. A (Hamel) basis, by definition should generate every element as a (finite, by definition) linear combination of the basis. A Hilbert space basis only generates, by finite combinations, a dense subspace, and then you can get the remaining elements as infinite sums. A separable Hilbert space has a countable Hilbert space basis, but not a countable vector space (Hamel) basis.
– user119256
Jan 14, 2014 at 17:58

Definition: A vector space basis (or Hamel basis) is a system of vectors such that every vector of the space if a linear combination (by definition only finitely many of the coefficients are non-zero) of the vectors in the system.

Examples:

1. The bases you know of finite dimensional vector spaces.
2. $\{1,x,x^2,...\}$ basis of $\mathbb{R}[x]$
3. Hamel's basis of $\mathbb{R}$ as a $\mathbb{Q}$-vector space.

Non-example:

1. $\{1, \sin(nx),\cos(nx)\}_{n=1}^{\infty}$ as a system of vectors of $L^{2}([0,2\pi])$. Because the function with graph the picture below (multiply the numbers in the picture by $\pi$) is not a finite combination of these elements. In particular a finite combination of these must be continuous but the graph below is not.

Definition: A Hilbert-space-orthogonal-basis (I put it with dashes to emphasize it is a concept by itself and not the union of the union of its parts) is an orthogonal system that is complete, where complete means that its linear combinations (finite as before) are dense in the space.

Example: The non-example above is an example of Hilbert-space-orthogonal-basis.

It is a theorem that when you have an a system of vectors that is an orthogonal-basis-of-a-Hilbert-space you can express every vector of the space as infinite linear combinations (Fourier series) of this system.

Theorem: A Hilbert space with a countable Hilbert-space-orthogonal-basis (what is called a separable Hilbert space) can only have an uncountable vector space basis (a Hamel basis).

Proof: Assume $H$ is Hilbert and $B=\{v_1,v_2,...\}\subset H$ is countable. Let $E_n$ be the set of all linear combinations of $\{v_1,v_2,...,v_n\}$. Since $E_n$ is finite dimensional it is closed in $H$. Since $H$ is infinite dimensional $E_n\neq H$. Therefore $E_n$ cannot contain any ball of $H$ (it is nowhere dense). But if $B$ is a basis then $H=\cup_n E_n$. Because $H$ is Hilbert, it is a complete metric space. By Baire category theorem it cannot be a union $\cup_n E_n$ of the countably many nowhere dense sets $E_n$.

Theorem: All Hilbert-space-orthogonal-bases have the same size.

Proof: Assume $B$ is a countable Hilbert-space-orthogonal-basis of the Hilbert space $H$ and $A$ is another Hilbert-space-orthogonal-basis of $H$. Expand each vector of $B$ as an infinite linear combination (Fourier series) of $A$. If $b\in B$ is given as $$b=\sum a_i v_i,$$

with $a_i$ scalars and $v_i\in A$. Then, by Parceval's identity $$||b||^2=\sum |a_i|^2||v_i||^2.$$

Being this a convergent series of reals, only countably many can be non-zero. So, the countably many elements of $B$ are generated, as infinite combinations, (each) by countably many elements of $A$. In total only countably many elements of $A$, say $C\subset A$, are needed to generate all elements of $B$. Therefore, if $v\in A\setminus C$, then $v$ being orthogonal to all elements of $C$ would be orthogonal to all elements of $B$. Therefore $v=0$.

Conclusion: An infinite dimensional separable Hilbert space (although it has a countable Hilbert-space-orthogonal-basis) it cannot have a vector space basis (a Hamel basis) that is also orthogonal.

Proof: From the theorems above.

1. An uncountable set cannot be orthogonal in a separable Hilbert space.
2. A vector space basis (a Hamel basis) of an infinite dimensional separable Hilbert space is uncountable.
• thank you so much for your detailed answer and your time. Okay i now understand the two notions of basis and Hilbert basis. You'r saying: 1-In infinite dimensional Hilbert space all vector space bases are uncountable 2- Those vector space bases are not orthogonal. But i am confused that you are still using the word 'seperable Hilbert space'. So do we need somethig like "a Hilbert space with uncountable vector space is non-seperable" or are we concluding that the answer of my problem is "No there isnt"? I hope i am not annoying Jan 15, 2014 at 7:29
• @iamvegan Separability is a topological notion. A topological space is separable when it has a countable dense set. When a Hilbert space has a countable Hilbert-space-orthogonal-basis, this generates a dense set. If we then take all linear combinations with rational coefficients of this basis we get a countable dense set. That is why Hilbert spaces with countable Hilbert-space-orthogonal-basis are called separable. Still their vector space bases are uncountable.
– user119256
Jan 15, 2014 at 12:59
• I think i've just had the light bulb moment, thanks Jan 15, 2014 at 13:23