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?


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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 '14 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. $\endgroup$ – Dave L. Renfro Apr 21 '14 at 21:28
  • $\begingroup$ Thanks, those are really helpful comments! $\endgroup$ – Marius Kempe Apr 22 '14 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? $\endgroup$ – echinodermata Apr 24 '14 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 '14 at 18:26

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... $\endgroup$ – Marius Kempe Apr 21 '14 at 7:31
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    $\begingroup$ There is no such thing as «being differentiable \omega times», really. $\endgroup$ – Mariano Suárez-Álvarez Apr 21 '14 at 7:36
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    $\begingroup$ Of course; I thought there might be a clever definition. $\endgroup$ – Marius Kempe Apr 21 '14 at 7:37

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