# Is there a basis which spans the real numbers?

Is there a finite set of real numbers $S=\{a_1, a_2, ..., a_n \}$ such that every real number can be written as a linear combination (with integer coefficients) of the elements of $S$? If no, is there a set $M= \{a_1, a_2, ... \}$ with a countably infinite number of elements, such that every real number can be written as a linear combination (with integer coefficients) of the elements of $M?$

I suspect the answer is no for both. However, if $S$ and/or $M$ exist, they must contain at least one transcendental number; a linear combination of only algebraic numbers should still be algebraic.

Motivation:

First, I solved a simpler problem. I wanted to see if we can span the integers with a finite basis containing any integers except $1$. The answer is yes. One such basis is $\{2, 3 \}$, because for any integer $n$, we have $n=n \cdot (3-2) = 3n - 2n=3n+2(-n).$

Next, I want to know if this can be done with rationals. I haven't worked on it, but I think I might have a shot at discovering the answer myself. However, I'm pretty sure that I can't solve the case for real numbers.

• A linear combination with which coeffients? Over $\mathbb R$ you just need 1. With coefficients in $\mathbb Q$ or $\mathbb N$ you can't have a finite basis! Jul 16, 2016 at 4:52
• Over what ring do you want your basis? Obviously $\mathbb{R}$ is one-dimensional over itself. And it is infinite-dimensional over $\mathbb{Q}$.
– ಠ_ಠ
Jul 16, 2016 at 4:54
• When talking about a basis, you usually want every element (i.e. real number) to be representable using a finite linear combination of basis elements. If your basis is countable, and your coefficient set is counrable, then the set of possible finite linear combinations is countable. Jul 16, 2016 at 5:04
• @Ovi: the answer to the question is no: the number of linear combinations of a countably infinite set with integer coefficients is countable, because the integers are countable. Jul 16, 2016 at 5:07
• A basis spans a vector space. A vectors space is taken over a field, Integers aren't a field. You shouldn't call it a basis then. Maybe I make mistake if so I apologize. Jul 16, 2016 at 7:20

This is essentially addressed in the comments, but let me flesh it out (if only to move this question off the "unanswered" queue):

Your suspicion that this isn't possible is correct, and the easiest way to see this is via cardinality.

If $$A$$ is any countable set of reals and $$C$$ is any set of "allowed coefficients" (you use $$\mathbb{Z}$$, but $$\mathbb{Q}$$ would give the same answer, as would the set of algebraic numbers, or so on and so forth), then the set $$LinComb(A,C)=\{x: \exists c_1,..., c_n\in C, a_1,..., a_n\in A(x=c_1a_1+...+c_na_n)\}$$ is countable. Showing this is a good exercise; the key points are:

• The union of countably many countable sets is countable.

• The Cartesian product of two (hence, of finitely many) countable sets is countable.

• We can break $$LinComb(A,C)$$ into countably many "pieces" - namely, for each $$n\in\mathbb{N}$$ we have the set $$LC_n(A,C)$$ of reals which can be written as a linear combination of elements of $$A$$ using coefficients from $$C$$ of length $$n$$.

Now, what can you say about the cardinality of $$LC_n(A,C)$$, and how that determines the cardinality of $$LinComb(A,C)$$?

So $$LinComb(A,C)$$ is countable as long as $$A$$ and $$C$$ are. But the reals are uncountable, so we'll never have $$LinComb(A,C)=\mathbb{R}$$.

• Isn't your third bullet missing exactly the most important "piece", namely the linear combinations which, for every $n\in\mathbb{N}$, there exists a non-zero coefficient $c_m$ with $m>n$? Nov 29, 2022 at 20:59
• @RonKaminsky No, it's not missing anything. Infinite sums are not linear combinations (cf. the difference between "Hamel basis" and "Schauder basis"). Nov 29, 2022 at 21:04
• Thanks, terminology always gets me! I have to wonder, then, why this is tagged "calculus". Nov 29, 2022 at 21:38

$1$ is a basis which spans the real numbers. Also a vectors space to have the said basis requires coefficients from a field.

• I mentioned in the question that the coefficients must be integers
– Ovi
Jul 16, 2016 at 7:04
• Here seems to be a related question math.stackexchange.com/questions/286017/… Jul 16, 2016 at 7:06