# The dimension of the real continuous functions as a vector space over $\mathbb{R}$ is not countable?

This question is out of curiosity. I first attempted a web crawl for this answer but was befuddled when Google didn't turn up the result after a couple of tries. If anyone has a reference, I'd be appreciative.

The ultimate thing I want to know is whether or not $C(\mathbb{R},\mathbb{R})$ has countably-infinite dimension over $\mathbb{R}$. However, I have an inkling that $C([a,b],\mathbb{R})$ itself does not have countably-infinite dimension.

I attempted to show that if $\{f_n\}_{n=1}^\infty$ is a purported basis for $C([a,b])$ then $\sum_{n=1}^\infty 2^{-n}f_n$ is not in their span, but I quickly found out that I don't know how quickly any of these $f_n$ increase.

I would like a constructive proof if one can be given; however, I would begrudgingly accept an existence proof. I don't mind if you feel the need to add a metric or topologize the space in any way.

The functions $f_t(x) = e^{tx}$, $t \in \mathbf R$ are linearly independent over $\mathbf R$. Can you prove it?

(Hint: suppose you have a minimal linear dependence relation among them, and use its derivative to produce a combination that is yet minimaler.)

Remark: we don't need the axiom of choice for this.

• +1 for the user of "minimaler". :-) Commented Feb 5, 2014 at 2:35
• @Nicholas I thought it was the bestest word to use in this context. Commented Feb 5, 2014 at 2:38
• Yes, I got it. Thank you. I can show they are linearly independent, but I am confused by what you mean when you say 'minimal linear dependence'.
– user123641
Commented Feb 5, 2014 at 2:46
• You are welcome, @Bryan. I mean one where the number of nonzero coefficients is minimal Commented Feb 5, 2014 at 2:47

Indeed $C([a,b],{\mathbb R})$ does not have countably-infinite dimension: no infinite-dimensional Banach space does (e.g. by the Baire category theorem).

• I must say that I like this answer; Baire spaces are dear to my heart, but I haven't studied Banach spaces in-depth and Bruno's answer is more 'hands-on'. It's always good to have more than one explanation however. Thank you.
– user123641
Commented Feb 5, 2014 at 2:41