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?
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$\begingroup$ I'm hacking together a reference sheet so space is at a premium. :) $\endgroup$– badpCommented Jul 6, 2011 at 9:08
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7$\begingroup$ "The set A is not infinite". $\endgroup$– MarkCommented Jul 6, 2011 at 10:27
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5 Answers
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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.
answered Jul 6, 2011 at 8:53
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3$\begingroup$ Unless the "reader" is a computer, in which case a symbolic language is far more precise. $\endgroup$– LBushkinCommented Jul 7, 2011 at 3:30
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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 $\mathbb{N}^+ = \lbrace 1,2,3, \ldots \rbrace$).
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$\begingroup$ The empty set is finite, I believe. What does that mean $|X|<\mathbb N^+$ anyway? Do you mean $\in$? $\endgroup$– Asaf Karagila ♦Commented Jul 6, 2011 at 12:30
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$\begingroup$ @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. $\endgroup$ Commented Jul 6, 2011 at 12:47
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$\begingroup$ 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 :-) $\endgroup$– Asaf Karagila ♦Commented Jul 6, 2011 at 12:57
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Why not simply, "$A$ is finite" or $|A| = n$.
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5$\begingroup$ 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. $\endgroup$ Commented Jul 6, 2011 at 10:06
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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.
answered Jul 6, 2011 at 13:00
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$\begingroup$ I'm not clear about your use of $\emptyset$ in that formula. Did you mean $\emptyset\subsetneq S$? $\endgroup$– Asaf Karagila ♦Commented Jul 6, 2011 at 13:22
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$\begingroup$ To me $\subset$ is a strict, $\subseteq$ is nonstrict. $\endgroup$ Commented Aug 7, 2014 at 21:24
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$\begingroup$ I think I got it three years ago. :-P $\endgroup$– Asaf Karagila ♦Commented Aug 7, 2014 at 21:44
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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.
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2$\begingroup$ If $A$ is amorphous then there is no injection from $\mathbb N$ into $A$, but $A$ is most certainly not finite. $\endgroup$– Asaf Karagila ♦Commented Jul 6, 2011 at 12:29
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1$\begingroup$ I have never come across amorphous sets, but they seem to assume $\lnot AC$ in their definition. Which is silly. $\endgroup$– user1729Commented Jul 6, 2011 at 13:18
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$\begingroup$ What you <i>can</i> do is not always such a good idea. $\endgroup$ Commented Jul 6, 2011 at 13:18
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2$\begingroup$ Why is it silly not to assume $AC$? $\endgroup$– Asaf Karagila ♦Commented Jul 6, 2011 at 13:20
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$\begingroup$ ...you mean you are happy with only the one orange?... $\endgroup$– user1729Commented Jul 6, 2011 at 13:22