# Is every real a finite linear combination of a proper subset of the reals with rational coefficients?

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.

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

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.
• 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). Jul 10, 2015 at 4:19