2
$\begingroup$

Wolfram Mathworld explains a Hamel basis as "a set of real numbers $\left\{U_\alpha\right\}$ such that every real number $\beta$ has a unique representation of the form:

$$ \beta = \sum\limits_{i=1}^{n}r_i U_{\alpha_i} $$

where $r_i$ is rational and $n$ depends on $\beta$."

Am I correct to interpret this as meaning that every real number is a unique finite linear combination of real numbers from a proper subset of $\mathbb{R}$ with coefficients in $\mathbb{Q}$? This would be quite a fascinating result to me.

$\endgroup$
1
  • $\begingroup$ Is $\mathbb R $ as a V.S over $\mathbb Q$ countably- or uncountably infinite-dimensional? I think it is clearly uncountable. $\endgroup$
    – Gary.
    Jul 10, 2015 at 3:45

1 Answer 1

Reset to default
3
$\begingroup$

Yes, your understanding is correct, but it isn't so surprising. We could take the proper subset of $\Bbb R$ to be $\Bbb R\setminus \{0\}$. Then every real except $0$ is representable with $n=1$ and $0=1+(-1)$ Clearly $\{U_\alpha \}$ is uncountable. Otherwise you could invoke the countable union of countable sets to show the set of sums is countable, which the reals are not. If the set of $\{U_\alpha \}$ is uncountable, all bets are off.

$\endgroup$
2
  • 2
    $\begingroup$ This is indeed an example of a proper subset that spans ${\mathbb R}$ (in the sense that every real number is a finite rational linear combination of elements from this set) but it is of course not an example of a Hamel basis (because the representations are not at all unique). $\endgroup$
    – WillO
    Jul 10, 2015 at 4:19
  • 2
    $\begingroup$ @WillO: Correct. It responds to the question as asked and I thought it would help OP to see why it was not surprising. Now to construct a basis we can build up-if there is a vector not represented, add it, or we can build down-if there is a vector represented more than once, delete one of the ones that represent it. When we are done either way, we have a Hamel basis. $\endgroup$
    – Ross Millikan
    Jul 10, 2015 at 4:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .