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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?

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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. –  user2468 Jul 29 '12 at 23:45

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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!

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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 0:23

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).

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