A set $S\subseteq\mathbb R$ is said to be linearly independent if for distinct $x_1,\ldots,x_k$ ($k\in\mathbb N$) and for integers $n_1,\ldots,n_k$, $$n_1x_1+\ldots+ n_kx_k=0$$ implies that $$n_1=\ldots=n_k=0.$$ It is not difficult to see that this definition is equivalent to the one in which $n_1,\ldots,n_k$ are allowed to be rationals.

By Zorn’s lemma, there exists a maximal such linearly independent $S$, which is a Hamel basis for $\mathbb R$ over $\mathbb Q$. Now, Problem 14.7 in Billingsley’s Probability and Measure (1995) claims that

[E]ach real $x$ can be written uniquely as $x=n_1x_1+\cdots+n_kx_k$ for distinct points $x_i$ in $S$ and integers $n_i$. [emphasis added]

I think this is not right:

(1) In the above definition of linear independence, integers and rationals are interchangeable, but... (2) ...in the Hamel-basis representation, rationals must be allowed. Integer coordinates are not enough to represent all of $\mathbb R$ for any maximal linear independent $S$.

I wonder if someone could confirm this is a typo. Thank you.

Here is a proof sketch for why integers are not sufficient. Let $x\in S$. Now, if Billingsley’s claim were true, it would be the case that $$\frac{x}{2}=\sum_{i=1}^kn_ix_i$$ for distinct $x_i\in S$ and integers $n_i$. But then $$x-2n_1x_1-\ldots-2n_kx_k=0.$$ Because of linear independence, $x$ must coincide with some other $x_i$. Since the $x_i$’s are distinct, there must be precisely one such $x_i$. Then, by matching the coefficients, it must be the case that $$1-2n_i=0,$$ or $n_i=1/2$, which is not an integer.


1 Answer 1


Yes, the coefficients in the vector representation of a real number $x$ over $\Bbb Q$, must be rational. See: http://mathworld.wolfram.com/HamelBasis.html.


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