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I've seen the notation $C^\omega$ used for the set of real analytic functions (e.g. on an interval). Where does it come from? What exactly does it mean? What is the reason behind it? Who first used it?

Thanks!

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    $\begingroup$ $\omega$ generally means 'a little bit more than $\infty$'. Of course that's subjective, but it creeps up often. For example $S_\infty$ represents the permutations of the set $\Bbb N$ with finite support (every element only permutes a finite number of elements), but $S_\omega$ represents all of the permutations. Similar for $A_\infty$ and $A_\omega$ (the even permutations). $\endgroup$
    – user123641
    Apr 21, 2014 at 18:38
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    $\begingroup$ I'm not sure, but I believe the $C^n$ notation is due to Gilbert Ames Bliss. At least, you can find it defined in a footnote on the first page of his 1904 Trans. Amer. Math. Soc. paper Sufficient condition for a minimum with respect to one-sided variations. I don't know about $C^{\infty},$ but I suspect this came within 10 to 15 years. However, I doubt $C^{\omega}$ showed up until after World War 2 (maybe Bourbaki?), as I don't recall seeing it except in relatively recent literature. $\endgroup$ Apr 21, 2014 at 21:28
  • $\begingroup$ Thanks, those are really helpful comments! $\endgroup$ Apr 22, 2014 at 2:02
  • $\begingroup$ @Bryan that's the first time I've seen $S_\infty$ and $S_\omega$ used like that! Can you point to a reference for the notation? $\endgroup$ Apr 24, 2014 at 5:29
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    $\begingroup$ @echinodermata I couldn't find the source from whence I saw that usage when you asked the question, and I still haven't. But the pattern does occur elsewhere. For example, in his book Topology, Munkres uses the notaion $\Bbb R^\infty$ for real sequences that are eventually zero and uses $\Bbb R^\omega$ for all real sequences. $\endgroup$
    – user123641
    Jun 14, 2014 at 18:26

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The idea of the notation is that $\omega$ is the ordinal following all the finite ordinals. It is larger than all of them: being analytic is a bit more than being differentiable to all orders.

I doubt one can trace who first used the notation, really. (Is it at all interesting?)

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    $\begingroup$ So there's no theorem behind it? I was hoping perhaps that a function that is differentiable $\omega$ times is analytic or something like that... $\endgroup$ Apr 21, 2014 at 7:31
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    $\begingroup$ There is no such thing as «being differentiable \omega times», really. $\endgroup$ Apr 21, 2014 at 7:36
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    $\begingroup$ Of course; I thought there might be a clever definition. $\endgroup$ Apr 21, 2014 at 7:37

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