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I've never seen phrases like "$\sqrt{5}$ people" or "a set with $\pi$ many elements". Are there sets with cardinality, say, $\frac{1}{2}$?

Edit: As Brian M. Scott pointed out, the only real numbers that are cardinalities of sets are the non-negative integers. Could you explain why is it so?

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    $\begingroup$ As the term cardinality is normally used, the only real numbers that are cardinalities of sets are the non-negative integers. The cardinalities of infinite sets, of course, are not natural numbers. $\endgroup$ – Brian M. Scott May 28 '13 at 17:55
  • $\begingroup$ @Ali: Yes, but that really doesn’t have anything to do with the question. $\endgroup$ – Brian M. Scott May 28 '13 at 18:02
  • $\begingroup$ I see my mistake, I'll remove my comment. $\endgroup$ – Ali May 28 '13 at 18:03
  • $\begingroup$ @Brian Alright, but why is that so? What is so special about natural numbers in this sense? $\endgroup$ – Xena May 28 '13 at 18:05
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    $\begingroup$ @MJD I don't mean the cardinality of the set of options. I mean that the combinatorial game $n$ can be seen as the game where one player can do $n$ "free" moves when $n$ is a positive integer. I know this is not the same (it was just the first thing I thought of when a set with a non-integer cardinality was mentioned, since I always found it funny that one could have a game where one player had half a free move). $\endgroup$ – Tobias Kildetoft May 28 '13 at 19:02
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Fundamentally, cardinals and real numbers are different things. You can think of the cardinality of a set as some abstract object ("cardinal") assigned to it, in such a way that two sets get assigned the same cardinal if and only if there is a bijection between them.

But there is a natural way to identify the finite cardinals with the natural numbers: namely, identify a cardinal $a$ with the natural number $n_a$ such that the set $\{1, 2, \dots, n_a\}$ has cardinality $a$ (i.e. any other set with cardinality $a$ has a bijection with $\{1,2,\dots, n_a\}$ and any other set with cardinality $a$). (In this discussion, "natural numbers" includes 0.) This is a nice identification because it makes cardinal arithmetic match up with the arithmetic of natural numbers. For instance:

  • Ordering: $n_a \le n_b$ if and only if any set of cardinality $a$ has an injection into any set of cardinality $b$

  • Addition: If $A,B$ are disjoint and have cardinalities $a,b$ respectively, then the cardinality $c$ of their union $C = A \cup B$ has $n_c = n_a + n_b$.

  • Multiplication: If $A,B$ have cardinalities $a,b$ and $C = A \times B$ has cardinality $c$, then $n_c = n_a n_b$.

  • Exponents: If $A,B$ have cardinalities $a,b$, and $C = A^B$ is the set of all functions from $B$ to $A$, then $n_c = n_a^{n_b}$.

One could imagine a system that identified other (infinite) cardinals with real numbers other than the natural numbers. For instance, nobody could stop me from proposing a system in which the cardinality $\aleph_0$ of the set of integers is identified with the real number $22/7$. But this system wouldn't have the properties listed above. For instance, $22/7 \le 4$, but there is no injection from $\mathbb{Z}$ to $\{1,2,3,4\}$, and $\mathbb{Z} \times \{1,2,3,4,5,6,7\}$ definitely does not have the same cardinality as $\{1,2,\dots, 21,22\}$. In fact, it's not hard to see that there would be no way of identifying infinite cardinals with real numbers that would preserve the above list of properties.

The properties above are very useful, and so in order to preserve them, we generally do not attempt to identify any other real numbers with cardinals. In principle, there could be another system that, although it didn't satisfy the above properties, had some other useful properties. But I've never heard of one that was useful enough to attract much attention.

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  • $\begingroup$ Yes! Very correct!! There are absolutely no real numbers which represent cardinality. Using this is a huge source for later problems when talking about infinite sets, and comparing infinities, because many would mistake that $\infty$ from calculus is also some form of cardinality. $\endgroup$ – Asaf Karagila May 28 '13 at 20:00
  • $\begingroup$ (By the way, the real numbers can be embedded into the cardinals, at least order-wise, if one assumes there exists infinite Dedekind-finite sets. But not only then.) $\endgroup$ – Asaf Karagila May 28 '13 at 20:01
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I am copying Brian M. Scott's comment here, because I think it answers the question completely and correctly:

As the term cardinality is normally used, the only real numbers that are cardinalities of sets are the non-negative integers. The cardinalities of infinite sets, of course, are not natural numbers.

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