What is an infinite-dimensional space?

Infinitely dimensional spaces A space is infinitely dimensional, if it has no basis consisting of finitely many vectors. By Zorn Lemma (see here), every space has a basis, so an infinite dimensional space has a basis consisting of infinite number of vectors (sometimes even uncountable).

Is function space infinite-dimensional?

Given that the set of all functions on a finite set can be identified with a finite-dimensional Cartesian space, it’s hardly surprising the spaces of (smooth/continuous/measurable) functions on an interval are infinite-dimensional.

What is the dimension of function space?

The zero vector is given by the constant function sending everything to the zero vector in V. The space of all functions from X to V is commonly denoted VX. If X is finite and V is finite-dimensional then VX has dimension |X|(dimV), otherwise the space is infinite-dimensional (uncountably so if X is infinite).

Why are functions infinite-dimensional?

Since the powers of x, x0= 1, x1= x, x2, x3, etc. are easily shown to be independent, it follows that no finite collection of functions can span the whole space and so the “vector space of all functions” is infinite dimensional.

Which of the following is an infinite dimensional vector space?

The two examples I like are these: 1) R[x], the set of polynomials in x with real coefficients. This is infinite dimensional because {xn:n∈N} is an independent set, and in fact a basis. 2) C(R), the set of continuous real-valued functions on R.

Is L2 space infinite dimensional?

f(x)g(x) dx. With this structure, L2([0, 1]) is also an infinite dimensional Hilbert space. A Hilbert space always has an orthonormal basis, but it might be uncountable.

Is Hilbert space infinite dimensional?

Hilbert spaces arise naturally and frequently in mathematics and physics, typically as infinite-dimensional function spaces.

What is the dimension of a function?

In mathematics, the notion of an (exact) dimension function (also known as a gauge function) is a tool in the study of fractals and other subsets of metric spaces. Dimension functions are a generalisation of the simple “diameter to the dimension” power law used in the construction of s-dimensional Hausdorff measure.

Why is a function an infinite dimensional vector?

Then you enrich that formality by introducing ideas of limits and convergence. To use the Taylor series to represent a function you have to make sense of infinite sums. That goes well beyond vector space axioms.

What is a finite vector space?

Finite vector spaces Apart from the trivial case of a zero-dimensional space over any field, a vector space over a field F has a finite number of elements if and only if F is a finite field and the vector space has a finite dimension. The primary example of such a space is the coordinate space (Fq)n.

Are all dimensions infinite?

That all dimensions are infinite in length, and curved, in that infinity, to ‘wrap’, so that the infinite extension in one direction ‘wraps back’, to come from the infinite extension in the opposite direction.

Does the space of all functions have an infinite basis?

It does have an infinite basis: . It’s an infinite dimensional function space. Some other useful infinite dimensional function spaces are these: the space of all continuous functions, the space of all differentiable functions, and the space of a integrable functions.

Is the set of all polynomials of arbitrary degree an infinite function space?

The set of all polynomials of arbitrary degree is also a function space, but it doesn’t have a finite basis. It does have an infinite basis: [math]1,x,x^2,\\ldots,x^n,\\ldots [/math]. It’s an infinite dimensional function space.

Why is the space of all polynomials infinite dimensional?

The space of all polynomials (ie., vectors in integer powers of x) requires infinitely many basis vectors 1, x, x^2, … x^n, … so it is infinite dimensional. if you understand the question you understand the answer!

Is there an infinite set of linearly independent vectors?

(An infinite linearly independent set exists, since $X$ is infinite-dimensional. Normalizing the vectors does not influence the linear independence.) There is a Hamel basis $B$ containing this set. Then there is a linear function $\\Zobr fX{\\R}$ such that $f(x_n)=n$ and $f(b)=0$ for $b\\in B\\setminus\\{x_n; n\\in\\mathbb N\\}$.