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Upon looking at methods that show $\mathbb{R}$ is not finite dimensional over $\mathbb{Q}$ I came across a method mentioned here by the user Bill Dubuque, he took a set of vectors of the form $\log(p)$ where $p$ is prime and showed that the set is independent, but in his proof he only takes $n$-primes. So my questions are:

The set of these logarithms is infinite , why did he only use $n$ primes?

Why does this show that $\mathbb{R}$ is not finite dimensional?

For the second question I'm not sure but I think it is because no matter what $n$ is the set is independent, but I'm not sure this can be extended to $n= \infty$.

Community
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user10444
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    Because $\mathbb{R}$ (over $\mathbb{Q}$) has countably many linear independent vectors, namely $\log 2$, $\log 3$, $\log 5$, and so on. – Hanul Jeon Feb 03 '13 at 13:08
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    I think that the main point (as described in rschwieb's answer) was that in order to check the linear dependence of any set of vectors, you only need to check such linear combinations where there are only finitely many non-zero terms. There are two good reasons for this: A) this is the definition of linear (in)dependence, B) the linear combinations with infinitely many non-zero terms usually don't exist (except in some topological sense). – Jyrki Lahtonen Feb 03 '13 at 21:45

2 Answers2

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The method shows that a collection $\{\log p_i\ |\ i =1, \dots n\}$ is linearly independent and therefore $\dim_{\mathbb{Q}}\mathbb{R} \geq n$, but the argument works for any $n \in \mathbb{N}$, so $\mathbb{R}$ is an infinite-dimensional vector space over $\mathbb{Q}$.

Michael Albanese
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  • So in general to show the dimension is infinite its enough to show that it contains an independent set of any cardinality ? – user10444 Feb 03 '13 at 13:26
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    Yes. You have to be careful though. If you look at Arturo's answer to the same question, he discusses different sizes of infinity. If you want to show $\mathbb{R}$ is an infinite-dimensional vector space over $\mathbb{Q}$, then it is enough to show that there is a $\mathbb{Q}$-linearly independent set of any finite cardinality. This does not show however that $\mathbb{R}$ is an uncountably infinite-dimensional vector space over $\mathbb{Q}$. – Michael Albanese Feb 03 '13 at 13:35
  • @user10444 Well, what Gone wrote is actually a direct proof that ${\log p_i \mid i\in \Bbb N}$ is a linearly independent set over $\Bbb Q$. It shows that any nontrivial linear combination from the set produces something that's not zero. There is not really a need to argue with finite sets of LI vectors of every size, but that is correct too. – rschwieb Feb 03 '13 at 13:53
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    @user10444: Not "any cardinality" but "any finite cardinality". There are many more cardinalities which are infinite. – Asaf Karagila Feb 03 '13 at 13:57
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To show that $\{\log p_i\mid i\in \Bbb N\}$ is linearly independent over $\Bbb Q$, you would just need to show that every nontrivial linear combination of its elements is nonzero.

So, select finitely many elements of the set to combine. This set must have a highest index $\log p_i$, so without loss of generality, we can add all the $\log p_j$ into this linear combination that are absent simply by giving them the coefficient zero. Thus whatever finite linear combination you start with, you can always change it to include the first $n$ prime-logs.

The conclusion is that this combination (equal to the original combination) is nonzero, and so $\{\log p_i\mid i\in \Bbb N\}$ is LI over $\Bbb Q$.

rschwieb
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    The argument saying that there are LI sets of arbitrarily large finite size is fine too, but I thought this direct argument was worth pointing out too. – rschwieb Feb 03 '13 at 14:06