A simple question of finite number of basis

Question

Let $V$ be a vector space. Define

" A set $\beta$ is a basis of $V $" as "(1) $\beta$ is linearly independent set, and (2) $\beta$ spans $V$ "

On this definition, I want to show that "if $V$ has a basis (call it $\beta$) then $\beta$ is a finite set." In my definition, I have no assumption of the finiteness of the set $\beta$. But Can I show this statement by using some properties of a vector space?

Is $V$ a finite-dimensional vector space? If so, then prove that all basis have the same dimension. Then size of $β$ is the same as the standard basis: the $n$ column vectors of the $n×n$ identity matrix. — , Jul 29 '12 at 23:45

score 3 · Answer 1 · answered Jul 29 '12 at 23:39

3

Vector space does not need to have finite basis. For example look at the separable Hilbert space $L^2$ that has countable basis. Using axiom of choice you can show that each vector space admits basis (but there is no restriction on its size).

answered Jul 29 '12 at 23:39

dmm

788

score 1 · Accepted Answer · answered Jul 29 '12 at 23:41

1

I take it $V$ is finite-dimensional? For instance, the (real) vector space of polynomials over $\mathbb{R}$ is infinite-dimensional $-$ in this space, no basis is finite. And for finite-dimensional spaces, it's worth noting that the dimension of a vector space $V$ is defined as being the size of a basis of $V$, so if $V$ is finite-dimensional then any basis for $V$ is automatically finite, by definition!

answered Jul 29 '12 at 23:41

Clive Newstead

64,925

1

What you say is true, but: it is also common to define (rather early in the game) a vector space to be finite-dimensional if it has a finite spanning set. Then the fact that every finite-dimensional vector space has a finite basis is a nontrivial statement, whose proof (e.g. by row-reduction) can be found in any introductory linear algebra text. – Pete L. Clark Jul 30 '12 at 00:23

A simple question of finite number of basis

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