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The dimension of this vector space is obviously infinite dimensional, and it's not too much work to show that its basis is an uncountable set, making it an uncountably-infinite dimensional vector space.
Many questions have been asked on MSE along those lines.

My question is: for the vector space of all real-valued functions on $\mathbb{R}$, what is the cardinality of its basis set? If the basis has the same cardinality as $\mathbb{R}$ (which is $\aleph_1$, right?), that would be the dimension of the vector space, but it also might be a higher cardinal. How would one go about showing this formally?

Also, will anything change if the vector space is instead all real-valued functions on $[0,1]$? I see that vector space a lot but I don't think its size would be any different. Am I correct in thinking this?

purple_hat
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A set of vectors of cardinality $\alpha$ generates $\alpha\cdot c$ elements over $\Bbb R$ where $c=|\Bbb R|=2^{\aleph_0}$ (which equals to $\aleph_1$ only if the continuum hypothesis is assumed).

Since $\alpha\cdot\beta=\max(\alpha,\beta)$ for infinite cardinals $\alpha,\beta$, if a real vector space has cardinality $\beta>c$, then its dimension is also $\beta$.

Consequently, $\dim\Bbb R^{\Bbb R}=\dim\Bbb R^{[0,1]}=c^c=2^{\aleph_0\cdot c}=2^c$.

Berci
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  • This is helpful. But it seems to assume that there are $c$ real-valued functions on $\mathbb{R}$. How did you get that the vector space is $\mathbb{R}^\mathbb{R}$? – purple_hat Sep 07 '20 at 17:53
  • No, there are $c^c$ real functions. $A^B$ denotes the set of all functions $B\to A$. – Berci Sep 07 '20 at 17:55