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If I want to say that a set $A$ is numerable but infinite, I can do so like this: $$|A| = \aleph_0$$

What should I use instead to say that a set is finite? $|A|\in\mathbb{N}$? $|A|< \infty$? $|A|< \aleph_0$? Something else entirely?

Pete L. Clark
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badp
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5 Answers5

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You should say "The set $A$ is finite." There is nothing wrong with using sentences in mathematics; they often are easier for the reader to understand than a sequence of symbols.

Zev Chonoles
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7

In light of your comment below the question (in addition to "What should I use instead to say that a set is finite?"), I suggest using $|A| < |\mathbb{N}|$ (or $|A| < \aleph _0$); see Wikipedia's definition here and Theorem (5.4) here. Note that this allows $A$ to be empty (the empty set is finite, and has a cardinality of zero).

EDIT: Exact quotations from the above links: 1) "Any set $X$ with cardinality less than that of the natural numbers, or $|X| < |\mathbf{N}|$, is said to be a finite set" (where $\mathbf{N}=\lbrace 0,1,2,3,\ldots\rbrace $); 2) "A set $X$ is finite if and only if $|X| < |\mathbb{N}^+|$" (where $\mathbb{N}^+ = \lbrace 1,2,3, \ldots \rbrace$).

Shai Covo
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  • The empty set is finite, I believe. What does that mean $|X|<\mathbb N^+$ anyway? Do you mean $\in$? – Asaf Karagila Jul 06 '11 at 12:30
  • @Asaf: I meant to write $|X| < |\mathbb{N}^+|$ (as in that link); I'll fix this. Concerning the comment about the empty set, I see no problem with what I wrote. – Shai Covo Jul 06 '11 at 12:47
  • Very well. If $|X|<|\mathbb N^+|$ then it holds for the empty set, I was thinking it was meant to be $\in$ which would then imply the empty set is not finite :-) – Asaf Karagila Jul 06 '11 at 12:57
  • @Asaf: Thanks for this reply. – Shai Covo Jul 06 '11 at 13:01
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Why not simply, "$A$ is finite" or $|A| = n$.

Jack Rousseau
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    While "A is finite" is fine, "$|A|=n$" makes me ask What is $n$?? ($n=\aleph_0$?) You'll have to write "$|A|\in\mathbb N$" as in Swlabr's answer. – Hendrik Vogt Jul 06 '11 at 10:06
1

You could say,

"There does not exist an injection from $\mathbb{N}$ to $A$."

Personally, however, I would just go with either "$A$ is finite" or "$|A| <\infty$". It does depend on the context though.

user1729
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    If $A$ is amorphous then there is no injection from $\mathbb N$ into $A$, but $A$ is most certainly not finite. – Asaf Karagila Jul 06 '11 at 12:29
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    I have never come across amorphous sets, but they seem to assume $\lnot AC$ in their definition. Which is silly. – user1729 Jul 06 '11 at 13:18
  • What you can do is not always such a good idea. – ncmathsadist Jul 06 '11 at 13:18
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    Why is it silly not to assume $AC$? – Asaf Karagila Jul 06 '11 at 13:20
  • ...you mean you are happy with only the one orange?... – user1729 Jul 06 '11 at 13:22
  • (The `it is silly' was just my initial bias, but the more I think about it the more odd I believe it is silly. I mean, AOC is independent of ZF, so one can prove stuff in ZF and then it should holds in ZFC...but amorphous sets seem to go against this. There exist things in ZF which do not exist in ZFC. This is, I believe, silly...unless I am missing something?) – user1729 Jul 06 '11 at 14:04
  • Yes, you are missing the point that there are many things which are very interesting and only happen without the axiom of choice. In particular amorphous sets. I do not see why giving a definition which does not hold without the axiom of choice is a good idea when there are cleaner definitions which hold just fine in $ZF$. – Asaf Karagila Jul 06 '11 at 14:36
  • I understand what you are saying, I just don't understand how it can be true if AC is independent of ZF. I was meaning, unless I am missing something from this independence... – user1729 Jul 07 '11 at 09:11
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As a matter of style I would say, "$A$ is finite." It is best to avoid having a bristling obstacle course of symbols for your reader to penetrate. Which is easier to read here?

Every nonvoid subset of the positive integers has a least element.

$\forall \emptyset \subset S\subseteq {\Bbb N}$, $ \exists m\in S$ such that $m\le s$ $\forall s\in S.$

The choice is clear to me. Use notation and symbols to simplify and clarify.

ncmathsadist
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