# The name $\mathcal{C}^\omega$

Let A be a set (of real numbers); define $\mathcal{C}^\omega (A)$ as the set of all real-valued functions that are defined, bounded, and analytic on A.

My question is simply this: how did $\mathcal{C}^\omega (A)$ get its name? Who named it? What does $\omega$ mean in this context? How does it extend the notion of $\mathcal{C}^n (A)$, where $n \in \mathbb{N} \cup \{0, \infty \}$?

• I suspect you'll get a better answer by posting this on History of Science and Mathematics Stack Exchange. – Milo Brandt Dec 10 '15 at 5:19
• This started off being denoted $\mathrm{Anal}(X)$ but was changed for obvious reasons. (I jest.) – goblin Dec 10 '15 at 5:20
• @goblin On an unrelated note, one of my physics professors was complaining about functional analysis, saying it was useless and stupid etc. I had my functional analysis book with me and I showed it to him after class. He covered up "ysis" and told me "this is what I think of it". I chuckle whenever I think of it. – user223391 Dec 10 '15 at 5:22
• @avid19, nice :) – goblin Dec 10 '15 at 5:23
• @user173897, me too, but not as an innuendo, just as an abbreviation that accidentally turns out to be dirty (every time!). FWIW, I think we should probably refrain from posting any more comments about this silly little joke so not to make anyone feel uncomfortable. This is a math website, after all. – goblin Dec 10 '15 at 5:43

$\omega$ here is a symbol from logic corresponds to the ordinal number $\omega:=\cup_{n\to\infty} n$. $C^\omega$ is "more nice" as $C^\infty$, which I guess is what they wanted to get across and using another symbol for infinity seemed like a good way.

• $C^\infty$ means something else. – Andrés E. Caicedo Dec 10 '15 at 5:06
• Okay, so that was my thought. In that case, I have further questions (my understanding of ordinals is scant, so bare with me): As I understand it, $\mathcal{C}^\infty$ is a limit. And ordinals are not limits. I see then that they are sort of different. But, how does $\omega$ make it mean "analytic" (why is that the next step of all possible next steps)? How does one jump from the limit to the infinite ordinal? What is special about analytic functions here- why is there not an intermediate property between smoothness and analyticity? Can there be C^(omega+1), etc.? – user173897 Dec 10 '15 at 5:11
• It's important to note that not all smooth functions are analytic. – Antonios-Alexandros Robotis Dec 10 '15 at 5:12
• Smooth and analytic are two different but highly overlapping properties. I should have said "more nice" rather than "more smooth" in my answer. – Stella Biderman Dec 10 '15 at 5:14
• Ordinals are defined by limits. $\omega:=\lim_{n\to\infty} \cup_{i=0}^n i$. There are intermediate sets, the question is if there are ones that are important enough to name – Stella Biderman Dec 10 '15 at 5:17

You ask two questions:

1. Historically, how did this notation arise?

2. Logically, what is the reasoning behind it?

I will answer Q2 only.

The reasoning is that we are meant to think of $\omega$ as being a number larger than $\infty$, as in:

$$0 \leq 1\leq 2\leq\cdots \infty\leq \omega$$

The nice thing about this is that it gives us the inclusions in the correct order:

$$\mathcal{C}^0(X) \supseteq \mathcal{C}^1(X) \supseteq \mathcal{C}^2(X)\cdots \mathcal{C}^\infty(X) \supseteq \mathcal{C}^\omega(X)$$

But honestly, I find this notation fairly lame. For starters, it should really be the other way around, with $\infty$ being the biggest element and $\omega$ the second-biggest. But that's just the start of the problems.

If you think about it for awhile, it becomes clear that what we should do is define an altogether new sequence $\mathcal{A}$ as follows: $\mathcal{A}^n(X)$ consists of all real-valued functions $f$ on $X$ such that for all $x \in X,$ there exists a neighbourhood of $x$ on which the function $f$ equals its own $n$th-order Taylor polynomial about $x$. So basically, $\mathcal{A}^n(X)$ consists of all functions on $X$ that piecewise (on each connected component of $X$) expressible as polynomials of degree $n$ or less (including the $0$ polynomial).

It therefore makes sense to write $\mathcal{A}^\infty(X)$ for the analytic functions on $X$.

With these conventions, we have the following system of inclusions:

$$\mathcal{C}^0(X) \supseteq \mathcal{C}^1(X) \supseteq \mathcal{C}^2(X)\cdots \mathcal{C}^\infty(X) \supseteq \mathcal{A}^\infty(X) \cdots \supseteq \mathcal{A}^2(X) \supseteq \mathcal{A}^1(X) \supseteq \mathcal{A}^0(X)$$

Of course, all the subscripts should really be at the bottom, as in $\mathcal{C}_n$ and $\mathcal{A}_n$.

• I like all of the notation that you suggest! – user173897 Dec 11 '15 at 3:20
• @user173897, thanks. – goblin Dec 12 '15 at 2:19