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Is there a set $S\subset \Bbb{R}$ such that every $x\in\Bbb{R}$ can be writen as

$$x = a_1 s_1 + a_2 s_2 + \dots + a_n s_n$$

where $a_1, a_2, \dots, a_n\in\Bbb{Z}$ and $s_1, s_2, \dots, s_n\in S$, and $x = 0 \implies a_1 = a_2 = \dots = a_n = 0$?

In other words, is there a linearly independent spanning set for $\Bbb{R}$ with respect to $\Bbb{Z}$?

Alma Arjuna
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    There are only countably many elements in your span, so the answer (to the question as written) is no. – lulu Sep 29 '20 at 15:42
  • You need uncountably many $s_i$ ... A finite or countably infinite number will not work. – Ted Shifrin Sep 29 '20 at 15:52
  • @Arjuna196 A countable union of countable sets is again countable. See, e.g., this question. $\mathbb R$ is not countable, so your construction will not work. – lulu Sep 29 '20 at 15:54
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    @TedShifrin but I didn't say that $S$ is countably. – Alma Arjuna Sep 29 '20 at 15:54
  • Au contraire: You said you had $n$ elements, $s_1,\dots,s_n$. – Ted Shifrin Sep 29 '20 at 15:55
  • Perhaps your quantifiers were unclear. It looks as if you intended $S={s_i}_{i=1}^n$. – lulu Sep 29 '20 at 15:55
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    In this question, ${s_1, s_2, \dotsc, s_n}$ are clearly just some finite subset of $S$, just like in the usual definition of a linear combination of elements of a vector space. I think your question is asking "is $\Bbb R$ a free $\Bbb Z$-module?". The answer is no, because $\Bbb R$ is divisible. – Izaak van Dongen Sep 29 '20 at 15:56
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    I just said that $s_1, s_2, \dots, s_n$ are elements of $S$... – Alma Arjuna Sep 29 '20 at 15:56
  • I suggest editing your post for clarity. I expect many readers will assume that you intended $S$ to be finite when, in fact, you need it to be uncountable. Even for uncountable "bases" it is impossible, though, because (as others have remarked) $\mathbb R$ is divisible. – lulu Sep 29 '20 at 16:01
  • @IzaakvanDongen what does it mean to say that $\Bbb{R}$ is divisible? – Alma Arjuna Sep 29 '20 at 17:52
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    @Arjuna196, "$\Bbb R$ is divisible" means that for any $x \in \Bbb R$, $n \in \Bbb N$, we can solve the equation $ny = x$ for $y$. Basically, "you can divide by natural numbers". No free $\Bbb Z$-module can have this property, intuitively because in a free $\Bbb Z$-module, the elements are "column vectors" with integer components, which you clearly can't always divide by $n$. Wojowu's answer very nicely uses this concept to give a proper proof. – Izaak van Dongen Sep 29 '20 at 19:28

1 Answers1

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No, there is no such set $S$. If $S$ is linearly independent $s\in S$, then $s/2$ is not an integer linear combination of elements of $S$, as supposing we had $s/2=a_1s_1+\dots+a_ns_n$ for $s_i\in S$ and $a_i\in\mathbb Z$, we would find $2a_1s_1+\dots+2a_ns_n-s=0$. This linear combination is nontrivial since $s$ has an odd (hence nonzero) coefficient in this combination.

Wojowu
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  • I am now curious about the case of replacing $\Bbb{Z}$ with $\Bbb{Q}$. Should I make a new post with this question, or does it have any simple and immediate solution that would easily fit into a comment on this post? – Alma Arjuna Sep 30 '20 at 15:58
  • @Arjuna196 Every vector space over a field has a basis. – Wojowu Sep 30 '20 at 16:00
  • what would such a basis look like? – Alma Arjuna Sep 30 '20 at 17:05
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    @Arjuna196 Unfortunately an explicit description of such a basis is impossible. Its existence relies on the axiom of choice. – Wojowu Sep 30 '20 at 17:37