6
$\begingroup$

How would you find the cardinality of the basis of $\mathbb{R}$ over $\mathbb{Q}$? Is it countable or uncountable?

In general, how do you find the cardinality of a basis of an infinite-dimensional vector space? Do you just search for a bijection between the basis and, say, $\mathbb{N}$ or $\mathbb{R}$? What are some instructive examples?

$\endgroup$
  • 1
    $\begingroup$ A basis for $\mathbb{R}$ over $\mathbb{Q}$ is called a Hamel basis. $\endgroup$ – Mankind May 4 '15 at 22:19
6
$\begingroup$

If $V$ is a vector space over the field $F$, and $B$ is a basis for $V$ then every element in $v$ is the unique combination of finitely many elements from $B$ with coefficients from $F$.

So each $v\in V$ can be represented as a finite subset of $B\times F$ which is a function. How? $v(b)=\alpha$ if $b$ is a basis element which has a nonzero coefficient, $\alpha$, in the linear combination of $v$ from $B$.

How many such functions are there? Now let's put into play the assumption that $V$ is infinite.

At least $B\times (F\setminus\{0\})$, since each singleton $\{\langle b,\alpha\rangle\}$ corresponds to the vector $\alpha\cdot b$. But if $X$ is infinite, the set of all finite subsets of $X$ has the same cardinality as $X$. So we get a bijection using Cantor-Bernstein. And since we're talking about infinite sets, we can consider $F$ and not $F\setminus\{0\}$.

So we get that $|V|=|B\times F|=|B|\cdot|F|$. But multiplying two non-zero cardinals, at least one of them is infinite, we get that $|B|\cdot|F|=\max\{|B|,|F|\}$.

In particular, if $|V|>|F|$ then $|V|=|B|$, or in other words $\dim V=|V|$.

So now if we want to apply this to $\Bbb R$ as a vector space over $\Bbb Q$, what do we get?

$\endgroup$
  • $\begingroup$ In the second line, do you mean "coefficients from $F$"? And the second paragraph, do you mean "finite subset of $B \times F$? $\endgroup$ – user217664 May 5 '15 at 16:41
  • $\begingroup$ Thanks for catching that! (Yes to both.) $\endgroup$ – Asaf Karagila May 5 '15 at 16:50
  • $\begingroup$ Thank you, this was helpful. But how did you get $ |B \times F| = |B| \cdot |F| $? $\endgroup$ – user217664 May 5 '15 at 16:51
  • $\begingroup$ The definition of cardinal multiplication. $\endgroup$ – Asaf Karagila May 5 '15 at 16:53
0
$\begingroup$

The cardinality of a Hamel basis of $\mathbf R$ over $\mathbf Q$ can't be countable, since it would imply $\mathbf R$ is countable.

A more general result is that if $E$ is an infinite-dimensional Banach space over a subfield $K\subset\mathbf C$, Hamel bases have all cardinality $\lvert E\rvert$.

$\endgroup$
  • $\begingroup$ Of course I wasn't considering this trivial case. Thanks for pointing my formulation was ambiguous. It's corrected now. $\endgroup$ – Bernard May 4 '15 at 23:32
  • $\begingroup$ How can you have a Banach space over any field which is not complete (so $\Bbb R$ or $\Bbb C$)? Just take a Cauchy sequence of scalars which does not converge and use it to create a Cauchy sequence of vectors which does not converge. Moreover $\Bbb R$ is indeed a Banach space, over itself, so its dimension is $1$. As a vector space over $\Bbb Q$ it is not quite a Banach space anymore. So the "more general theorem" does not apply. Finally, indeed $\dim_\Bbb Q\Bbb R>\aleph_0$, but why does that mean it has to be equal to $2^{\aleph_0}$? $\endgroup$ – Asaf Karagila May 5 '15 at 4:25
  • $\begingroup$ A Banch space over a subfield $K$ is simply a complete normed space over $K$. It can be consided a vector space over $\mathbf R$ (or $\mathbf C$) extending the scalars by contonuity. However it can have a basis over $K$. See this paper for details. $\endgroup$ – Bernard May 5 '15 at 7:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy