Infinite dimensional vector spaces without basis?

[Aziza]

According to my professor, there exist infinite dimensional vector spaces without a basis, and he asked us to find one. But isn't this impossible? The definition of a dimension is the number of elements in the basis of the vector space. So if the space is infinite-dimensional, then the basis of that space has an infinite amount of elements..

the only vector space I can think of without a basis is the zero vector...but this is not infinite dimensional..

 

[pwsnafu]

Err, the statement "every vector space has a basis" is equivalent to the axiom of choice.

 

[Stephen Tashi]

According to my professor, there exist infinite dimensional vector spaces without a basis, and he asked us to find one. But isn't this impossible?

It's not impossible if your course materials say that every element in a vector space with a basis must be representable as a finite linear combination of the basis vectors or define the concept of "linear combination" as a finite sum. A basis for a vector space is usually defined that way. You'll have to read the fine print in how your materials define a vector space. It may be that "infinite dimensional" merely means "not finite dimensional" instead of implying that there is an infinite basis.

There are various modified definitions of a basis that are applicable to infinite dimensional vector space. See the "related notions" section of the Wikipedia article http://en.wikipedia.org/wiki/Basis_(linear_algebra).

 

[Fredrik]

As the two guys before me said, with the standard definitions, every vector space has a basis (assuming that we're working within the branch of mathematics defined by ZFC set theory).

Someone else asked about how to prove that every vector space has a basis a couple of weeks ago. This is what I said to him:

A basis is by definition a maximal linearly independent subset. This means that B is a basis of V if and only if B is linearly independent, and for all $x\in V-B$, $B\cup\{x\}$ is linearly dependent. Because of this, the existence of a basis in the finite dimensional case is actually trivial (if you understand the definitions perfectly).

A vector space V is said to be infinite dimensional if for all positive integers n, there's a linearly independent subset of V with cardinality n.

A vector space V that isn't infinite dimensional is said to be finite dimensional.

The dimension of a non-trivial (i.e. $\neq\{0\}$) finite-dimensional vector space V is defined as the largest integer n such that V has a linearly independent subset with cardinality n. This integer is denoted by dim V.

Theorem: Every non-trivial finite-dimensional vector space has a basis.

Proof: Let n be an arbitrary positive integer. Let V be an arbitrary vector space such that dim V=n. Let B be an arbitrary linearly independent subset of V with cardinality n. Clearly, for all $x\in V-B$, $B\cup\{x\}$ must be linearly dependent, because otherwise we would have dim V≥n+1>n.


The standard proof for the arbitrary case uses Zorn's lemma (which is equivalent to the axiom of choice). You will have to study some definitions to understand it. (In particular, the definition of "partially ordered set").

Theorem: Every non-trivial vector space has a basis.

Proof: Let V be an arbitrary non-trivial vector space. Let S be the set of all linearly independent subsets of V, partially ordered by inclusion. Let T be an arbitrary totally ordered subset of S. Clearly, $\bigcup T$ is an upper bound of T. Since every totally ordered subset has an upper bound, Zorn's lemma tells us that S has a maximal element.

 

[mathwonk]

we are saying the statement you ascribe to your professor is false in the presence if the usual axiom of choice, so you should verify what he said with him.

and it is easy to define an infinite dimensional space as one with no finite basis, without allowing a basis for the big one, and also the basis of {0} is the (finite) empty set.

 

[Vargo]

Perhaps he meant you to find a space without a countable basis.