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Is there a "standard" notation to denote the set of all finite subsets of $\mathbb{N}$? (or any set, not just $\mathbb{N}$)


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Doubt it. I would go with $\mathcal{P}_{\text{fin}}(\mathbb{N})$ or, depending on what exactly you're doing, just $S$. – Qiaochu Yuan Aug 4 '11 at 14:52
(some) people think of $2^{<\omega}$ (finite length binary strings) as the set of all finite subsets of $\omega.$ – jspecter Aug 4 '11 at 14:54
I think $2^{<\omega}$ is pretty ambiguous (what does a terminal $0$ mean?). The notation I see most often is some version of $[\omega]^{<\omega}$, with one or both occurrences of $\omega$ replaced by $\mathbb{N}$ or $\aleph_0$, depending on personal preference. Alternatively, I have also seen FIN, which has a nice blunt simplicity about it. – user83827 Aug 4 '11 at 15:07
@ccc: can you please add your reply as an answer so I can accept it – user10 Aug 4 '11 at 15:27
@user10: It looks like some more comprehensive answers have popped up in the meantime, so I'll just keep my comment as a comment unless you feel strongly otherwise. – user83827 Aug 4 '11 at 15:40
up vote 5 down vote accepted

Several possible notations for $\{A\subseteq\omega\mid |A|<\omega\}$:

  1. $[\omega]^{<\omega}$
  2. $P_\omega(\omega)$
  3. $\operatorname{Fin}(\omega)$

Where, of course, $\omega=\mathbb N$.

And as usual my advice on the matter: When in doubt, open with "We denote by [the chosen notation here] the set ..."

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Isn't $\omega$ the usual notation for the ordinal, rather than the cardinal? I'd be tempted to use $|A|\lt\aleph_0$ or $|A|\lt|\omega|$, rather than $|A|\lt\omega$... – Arturo Magidin Aug 4 '11 at 16:07
@Arturo: It is indeed confusing, but initial ordinals function as both ordinals and cardinals. In this case, $|\omega|=\omega$, so it is alright. I agree that it is somewhat confusing, and I do my best to use $\aleph_0$ when I only care about cardinality. The usual notations, however, use $\omega$ - and I do agree that it is less... cluttered than $\mathcal P_{\aleph_0}(\omega)$ or such. – Asaf Karagila Aug 4 '11 at 16:09
Yes, technically, cardinals are particular ordinals; but my impression is that one uses $\omega$ when one wants to emphasize/keep in mind the ordinal structure, and uses $|\omega|$ and $\aleph_0$ when one wants to ignore the ordinal structure. Clearly, my impression was mistaken, I'll have to try to track down where I got it from. – Arturo Magidin Aug 4 '11 at 16:14
@Arturo: Yes, in essence you are correct. However since one uses $\aleph$ notation a lot less than people would expect (as most of the time we use $\kappa,\lambda$ and such to denote cardinals) it is customary to just let them assume their "ordinal role" when needed, and to say an ordinal is of smaller cardinality than some initial ordinal is the same as saying it has a smaller order type. So it works out just fine. – Asaf Karagila Aug 4 '11 at 16:46
The second notation is rather confusing to me. If I saw it in a paper I would not be able to guess what it meant. – Qiaochu Yuan Aug 4 '11 at 16:52

You can find various notations, as mentioned in coments. (I doubt there is some generally accepted notation.)

  • You can find $[\omega]^{<\omega}$, e.g. here, which can be considered as a special case of $[A]^{<\kappa}$ - which denotes all subsets of $A$ of cardinality less then $\kappa$, see e.g. p.18 of the same book. In your case you could use $[\mathbb N]^{<\omega}$.

  • You can find $\mathrm{Fin}$, e.g. here and here

  • You can find $\mathbb N^{[<\infty]}$, e.g. here.

  • Hindman and Strauss use $\mathcal P_f(\mathbb N)$ in this book, which is similar to Qiaochu's suggestion $\mathcal P_{\mathrm{fin}}(\mathbb N)$.

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