1

What is the proper classification for the (infinite) dimensionality of $C^n$, the space of all functions (defined on $\mathbb{R}^m$, $m\in\mathbb{N}$) with continuous derivatives from order 0 to $n$?

I know this is a very broad (and possibly bad) question, but it came up in a problem and so I really would like to at least know where to look.

Arturo don Juan
  • 3,908
  • For some other nice answers check out this thread:

    http://math.stackexchange.com/questions/664084/the-dimension-of-the-real-continuous-functions-as-a-vector-space-over-mathbbr?rq=1

    In my answer below I've tried to flesh out Robert Israel's answer; Bruno Joyal's is also lovely but I suspect you'll find it more self-explanatory.

     – Pete Caradonna Jun 30 '16 at 10:48

1 Answers1

2

It's uncountable. There's a particularly nice proof of this using the Baire Category Theorem that I always was enamored by.

Assume that our infinite dimensional Banach space (and hence Baire space) $C^n$ has a countable basis $\{x_1, x_2, x_3, \ldots\}$ and define $C^n|_k = \textrm{span}\{x_1,x_2,\ldots, x_k\}$. Now by construction, $\cup_k C^n|_k = C^n$. But for all $k\in \mathbb{N}$ each $C^n|_k$ is closed and nowhere dense, and hence by the Baire Category Theorem we cannot have $\cup_k C^n|_k = C^n$, our contradiction!

Pete Caradonna
  • 3,292
  • Of course it's uncountable, but how uncountable is it (i.e. what kind of uncountable cardinality)? – Arturo don Juan Jul 17 '16 at 19:28
  • How could it conceivably be higher dimensionality than $|\mathbb{R}|= \frak{c}$ ? Your domains' cardinality is $\frak{c}$... – Pete Caradonna Jul 17 '16 at 20:24
  • With $n\geq 1$, let $f_1(x)=a_1$ and $f_2(x)=a_2 x$, where $a_{1,2}\in\mathbb{R}$. The dimensionality of the function space that those two span is $|\mathbb{R}\times \mathbb{R} |$, which is greater than just $\mathbb{R}$. Heck, I don't even know if all of the function space can be spanned by a countable amount of Cartesian products of 1-dimensional function spaces. – Arturo don Juan Jul 17 '16 at 20:51
  • Consider another angle: the cardinality of your function space of interest is also only continuum. – Pete Caradonna Jul 17 '16 at 22:09