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This question came up as a discussion I had with a friend. Clearly, the basis is not of finite cardinality since that would imply the set $\mathbb{R}$ has the cardinality $\aleph_{0}$ which is false. Therefore the basis clearly has to be infinite. However, the question that came up was that is the basis countably infinite or uncountably infinite. So I'd like to know the cardinality of the basis and also appreciate some pointers to the proof. Thanks in advance. :)

sayantankhan
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The size of the basis cannot exceed $\aleph$, as the set $\mathbb{R}$ spans your vector space.

The size of the basis is indeed $\aleph$. To see this, suppose we use a basis of size $\aleph_0$ (or any other cardinality less then $\aleph$ , for that matter). Let us denote this basis by ${x_1,x_2,x_3,...,x_n,...}$. Any element of $\mathbb{R}$ can be written as a finite linear combination of the form $\sum_{i=1}^{M} q_{i}x_{r_i}$ where $q_{i}\in\mathbb{Q}$.

In particular, if we fix some element $y\in\mathbb{R}$, all of the indices ${r_i}$ appearing in the linear combination are no more then some $N$, meaning that we can write $y$ as a linear combination of $x_1 ,..., x_N$.

Because $y\in\mathbb{R}$ was arbitrary, we conclude that: $$ \mathbb{R}\subseteq \bigcup_{N=1}^{\infty} span_\mathbb{Q} (x_1,...,x_N) $$

As you said yourself, each of the sets $span_\mathbb{Q} (x_1,...,x_N)$ is countable, meaning that the union $ \bigcup_{N=1}^{\infty} span_\mathbb{Q} (x_1,...,x_N)$ has cardinality $\aleph_0$, so it cannot contain $\mathbb{R}$.

Miel Sharf
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    Miel uses the (old-fashioned) notation $\aleph$ by itself for the cardinal of the continuum. – GEdgar Nov 18 '14 at 14:31
  • Thanks a lot. I realized I wasn't appreciating the point that an element has to a finite linear combination. Your answer really cleared stuff up. – sayantankhan Nov 18 '14 at 15:36
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For infinte dimensions, the cardinality of a vector space over $\mathbb Q$ equals the cardinality of the base.

Hagen von Eitzen
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Since any element fo $\;\Bbb R\;$ has to be a finite $\;\Bbb Q$- linear combination of some basis, the basis can not be countable, either. Thus...

Timbuc
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    Thus... It has to be one of the uncountable cardinals between $\aleph_0$ and $2^{\aleph_0}$... – Asaf Karagila Nov 18 '14 at 14:12
  • Indeed so, @AsafKaragila...and being a member of my Alma Mater and a student of logic and/or set theory, you surely know that accepting the (in)famous Continuum Hypothesis ,that one thus would be $;2^{\aleph_0};$ . Now, not accepting that would make things way more interesting...perhaps. – Timbuc Nov 18 '14 at 14:14
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    Yes, but why should this depend in the continuum hypothesis? It shouldn't, it doesn't. – Asaf Karagila Nov 18 '14 at 14:21
  • I think it does, @AsafKaragila: if the hypothesis is true then the dimension is for $;2^{\aleph_0};$ , otherwise...who knows? – Timbuc Nov 18 '14 at 14:23
  • Oh, I see: the dimension is the continuum=cardinality of the reals, never mind the hypothesis...right on! – Timbuc Nov 18 '14 at 14:24
  • I do, and everyone who would do the math will know as well. See the duplicate I suggested. – Asaf Karagila Nov 18 '14 at 14:24
  • Yes, so do I...without even seeing any duplicate. Thanks. – Timbuc Nov 18 '14 at 14:25
  • The axiom of choice is quite enough for this, no need for CH. – GEdgar Nov 18 '14 at 14:33
  • @GEdgar: Depending on how you state CH without choice, it might not be enough on its own! (E.g. it is possible that there is no intermediate cardinal, but there is no Hamel basis either!) – Asaf Karagila Nov 18 '14 at 14:35