# About the definition of vector space of infinite dimension

• I
• cianfa72
cianfa72
TL;DR Summary
About the definition of vector space in case of infinite dimension
Hi, a doubt about the definition of vector space.

Take for instance the set of polynomals defined on a field ##\mathbb R ## or ##\mathbb C##. One can define the sum of them and the product for a scalar, and check the axioms of vector space are actually fullfilled.

Now the point is: if one consider a sum of infinite polynomials (either countable or not) then the result might not be a polynomial at all !

I'm aware of the above "issue" boils down to the "completeness" of the metric space built over the vector space.

Does that means there is actually a "restriction" in the definition of vector space (like one must consider only finite sums of elements in the set) ?

Thank you.

Last edited:
A vector space by definition only has finite sums defined. I don't think I would call that a restriction, even in finite dimensions a vector space only has a single sum function which adds two elements, and by induction you can prove you can add any finite number of elements using that function. Nothing at any point tells you how to add infinite elements until you introduce a metric and start talking about notions of convergence.

PeroK and cianfa72
Office_Shredder said:
and by induction you can prove you can add any finite number of elements using that function. Nothing at any point tells you how to add infinite elements until you introduce a metric and start talking about notions of convergence.
Ah ok, by induction one can prove the sum is an element of the set only for a finite number of elements.

Another point related to this.

Consider for instance the Hilbert space of ##L^2(\mathbb R)## Lebesgue square-integrable functions on ##\mathbb R##. One can show it is separable i.e. there is a countable basis of orthonormal vectors/functions (these basis's elements are dense in ##L^2(\mathbb R)## according the topology given by the inner product's induced metric).

Now my point is: even though there is a countable basis, may the "cardinality" of the space itself be uncountably infinite ?

Last edited:
cianfa72 said:
Another point related to this.

Consider for instance the Hilbert space of ##L^2(\mathbb R)## Lebesgue square-integrable functions on ##\mathbb R##. One can show it is separable i.e. there is a countable basis of orthonormal vectors/functions (these basis's elements are dense in ##L^2(\mathbb R)## according the topology given by the inner product's induced metric).

Now my point is: even though there is a countable basis, may the "cardinality" of the space itself be uncountably infinite ?
##\mathbb{R}## is a one dimensional vector space over itself, yet the cardinality of that vector space is uncountable.

The more interesting question is how much bigger than the cardinality of the field is the cardinality of the vector space

I think my doubt is somehow related to the concept of Hamel/algebraic basis.

For example in ##L^2([a,b])## there is an orthonormal Hilbert basis dense in it, however it is not an Hamel basis since its linear span is not the entire space.

cianfa72 said:
I think my doubt is somehow related to the concept of Hamel/algebraic basis.

For example in ##L^2([a,b])## there is an orthonormal Hilbert basis dense in it, however it is not an Hamel basis since its linear span is not the entire space.
A hamel basis can generate exactly any element with a finite sum of basis elements. A Hilbert basis just has to be able to get arbitrarily close via finite sums. So Hilbert bases can often be smaller sets

I'm not sure what your doubt is so I don't know exactly how to address it...

cianfa72
Office_Shredder said:
A hamel basis can generate exactly any element with a finite sum of basis elements. A Hilbert basis just has to be able to get arbitrarily close via finite sums. So Hilbert bases can often be smaller sets
Yes, ok. The linear span of an Hilbert basis is, by definition, dense in the Hilbert space ##H##.

Any vector ##v## in ##H## can be given as countable linear combination of Hilbert basis's elements in the sense of convergent series to ##v## according the norm induced by inner product.

Does the above statement hold true in general also for non-separable Hilbert spaces?

Last edited:
cianfa72 said:
Yes, ok. The linear span of an Hilbert basis is, by definition, dense in the Hilbert space ##H##.

Any vector ##v## in ##H## can be given as countable linear combination of Hilbert basis's elements in the sense of convergent series to ##v## according the norm induced by inner product.

Does the above statement hold true in general also for non-separable Hilbert spaces?

Yes, every Hilbert space has an orthogonal basis. I think you need to assume the axiom of choice for this to be true or something.

cianfa72
Office_Shredder said:
Yes, every Hilbert space has an orthogonal basis.
You mean an orthonormal system of elements that is complete - i.e. an (orthonormal) Hilbert basis.

For non-separable Hilbert spaces there is not a countable Hilbert basis, however any vector ##v## in ##H## can be given as a convergent series of basis's elements.

cianfa72 said:
You mean an orthonormal system of elements that is complete - i.e. an (orthonormal) Hilbert basis.

For non-separable Hilbert spaces there is not a countable Hilbert basis, however any vector ##v## in ##H## can be given as a convergent series of basis's elements.

Yes

From this page any separable Hilbert space is isomorphic to ##\mathcal l^2##. I believe that holds true only for infinite dimensional separable Hilbert spaces (i.e. with a infinite countable orthonormal Hilbert basis), otherwise for finite dimensional Hilbert space over ##\mathbb C## it should be isomorphic to ##\mathbb C^n##.

Last edited:
cianfa72 said:
From this page any separable Hilbert space is isomorphic to ##\mathcal l^2##. I believe that holds true only for infinite dimensional separable Hilbert spaces (i.e. with a infinite countable orthonormal Hilbert basis), otherwise for finite dimensional Hilbert space over ##\mathbb C## it should be isomorphic to ##\mathbb C^n##.
What you said sounds right to me

cianfa72 said:
Now the point is: if one consider a sum of infinite polynomials (either countable or not) then the result might not be a polynomial at all !
Summation is a binary operation, hence you can extend it to only finite sums.

This ## \sum _{k=0}^\infty c_k x^k ## is not a polynomial.

The sum of two polynomials is a polynomial.

Does that means there is actually a "restriction" in the definition of vector space (like one must consider only finite sums of elements in the set) ?
That is indeed how vector spaces are defined. There is a generating set (of arbitrary cardinality) and the space is formed by considering all finite linear combinations of the generating set (over some fixed ground field, whether finite or infinite).

cianfa72 said:
Now my point is: even though there is a countable basis, may the "cardinality" of the space itself be uncountably infinite ?
Yes, because the ground field is uncountable. If you were to generate a space over ##\mathbb Q## using a countable generating set, the result would also be countable.

Last edited:
You can have additional structure in an infinite-dimensional vector space, such as a Schauder Basis , to allow for infinite sums . Such sums require a notion of convergence, which itself requires a topology.

Re completeness, every metric space is essentially as good as a complete one, in that it has a metric completion, i.e., it can be embedded densely in a complete metric space.

WWGD said:
You can have additional structure in an infinite-dimensional vector space, such as a Schauder Basis , to allow for infinite sums . Such sums require a notion of convergence, which itself requires a topology.
An infinite dimensional vector space has got, by definition, an algebraic/Hamel basis either infinite countable or uncountable.

By using Schauder basis (that has by definition countable cardinality), we can write any vector ##v## as a convergent series of its elements (convergent according the topology induced by metric if any).

In case of separable infinite dimensional Hilbert space ##H## there exists a countable orthonormal system of vectors/elements that is complete (an Hilbert basis). By definition the span of such Hilbert basis is dense in ##H##, hence it is a Schauder basis as well.

On the other hand, whether the infinite dimensional Hilbert space is not separable, there is not an Hilbert basis, yet it admits a Schauder basis I believe.

Last edited:
I think if a Hilbert space has a countable Schauder basis, then it has a countable Hilbert basis. You can basically just apply Grahm Schmidt on the Schauder basis and you get a set with the same span, and everything is orthonormal.

I suspect the same is true (cardinality of Schauder and hilbert bases are equal) in the non separable case but it's less obvious to me.

Existence of Schauder basis implies separable (e.g here)

cianfa72
nuuskur said:
Existence of Schauder basis implies separable (e.g here)
The inverse implication is straightforward: if an Hilbert space ##H## is separable then it has a countable complete orthonormal basis ##B##. Then any vector ##v## can be given as a convergent series in ##H## of elements in ##B##. Hence ##B## is a Schauder basis as well.

• Linear and Abstract Algebra
Replies
4
Views
1K
• Linear and Abstract Algebra
Replies
7
Views
574
• Linear and Abstract Algebra
Replies
8
Views
1K
• Linear and Abstract Algebra
Replies
8
Views
1K
• Linear and Abstract Algebra
Replies
5
Views
1K
• Calculus
Replies
11
Views
553
• Linear and Abstract Algebra
Replies
3
Views
1K
• Linear and Abstract Algebra
Replies
3
Views
3K
• Linear and Abstract Algebra
Replies
4
Views
1K
• Linear and Abstract Algebra
Replies
2
Views
1K