# An explicit basis for an infinite dimensional vector space.

Do you have any explicit example of an infinite dimensional vector space, with an explicit basis ?

Not an Hilbert basis but a family of linearly independent vectors which spans the space -any $x$ in the space is a finite linear sum of elements of the basis.

In general the existence of such a basis follows by the Axiom of choice but I wonder if there is at least one non trivial (not finite dimensional) case where we have some explicit constuction.

• Take the space of sequences of real numbers with finite support (only finitely many coordinates are non-zero), and the standard unit vectors. Feb 14, 2015 at 14:21
• Just as a note, there are set-theoretic arguments showing that you won't be able to do this for an infinite-dimensional Banach space. Feb 14, 2015 at 17:36

What about $$\beta=\{1, x, x^2, \ldots, x^n,\ldots \}$$

basis of $K[x]$ polynomial ring, where $K$ is any field. In this case we have $[K[x]: K] = \infty$.

Another example is

$$\gamma = \{1 , \cos x, \sin x, \ldots , \cos n x, \sin n x, \ldots \}$$

is a basis for the Euclidian Space $\mathscr{PC}[-\pi,\pi]$. It's used in Fourier series.

• The second one is a Schauder basis. The question asks for a Hamel basis. Feb 14, 2015 at 21:58

What do you think about $$X=\text{span}\{e_1,e_2,e_3,...\}$$

where $$e_i=(0,...,0,\underset{\underset{i^{th}\ place}{\uparrow}}{1},0,...).$$

• Maybe worth pointing out that this is the space of all sequences which are eventually zero. I believe this sequence space is often denoted $c_{00}$. Feb 14, 2015 at 16:39
• More generally, one could take the $e$'s to be members of any linearly independent set in a vector space (or take the span to be the formal sums of such elements). Feb 14, 2015 at 19:38

Rings of polynomials over any field in any number of indeterminates (viewed as vector spaces over the field) have the monomials as a basis.

Ring of integer-valued polynomials with coefficients in $\mathbf Q\,$ have as a basis:

$$\Bigl\{1, x, \frac{x(x-1)}2, \dots, \frac{x(x-1)\dotsm(x-k+1)}{k!},\dots\Bigr\}$$