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Can you ask given an infinite set about its cardinality? Does an infinite set have a cardinality? So, for example, what would be the cardinality of $+\infty$?

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Yes, it does. $+\infty$ is not a set, so it doesn't make sense to talk about its cardinality. –  Yury Dec 16 '12 at 19:55
Every set has a cardinality, but $+\infty$ is not a set. –  MJD Dec 16 '12 at 19:55
Philosophically speaking, it would make more sense to say that $+\infty$ has infinite magnitude rather than that it has infinite cardinality. Magnitude and cardinality are both notions of "size", but they are different. For example, the interval $[0,1]$ has finite magnitude but infinite cardinality. –  Trevor Wilson Dec 16 '12 at 20:06

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up vote 8 down vote accepted

Yes, infinite sets do have cardinality. First, however, note that cardinalities are not real numbers. In fact they are not natural numbers either. It just happens that the finite cardinalities coincide with the natural numbers (in representation as well in basic arithmetics). Therefore we don't really think about sets whose cardinality is "$+\infty$", in fact $+\infty$ is merely a formal notation for a quantity larger than all the real numbers. But cardinals are not real numbers, so we don't often denote cardinalities of infinite sets with $\infty$ (and when we do, it is often when we don't really care about the cardinality, just whether or not it is finite). Note that as a symbol $+\infty$ does not really make a set that we can measure its cardinality.

So infinite sets have cardinalities, the simplest one is $\aleph_0$ which denotes the cardinality of the natural numbers, i.e. $\{0,1,2,\ldots\}=\mathbb N$. True, it is not a "number" as most people would think about numbers, but numbers are simply mathematical objects representing a concrete quantity, and in this aspects $\aleph_0$ is no different.

We say that an infinite set $A$ has cardinality $\aleph_0$ if there is a function from $\mathbb N$ into $A$ which is a bijection. Not all infinite sets have cardinality $\aleph_0$. For example the real numbers have a strictly larger cardinality which can be computed as $2^{\aleph_0}$, the same cardinality as $\mathcal P(\mathbb N)=\{A\mid A\subseteq\mathbb N\}$.

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But... "$\infty$" isn't a set. A cardinality is the number of elements in a set. For example, the cardinality of $\{3,4,5\}$ is three.

Yes, infinite sets can also have cardinalities. The rigorous definition is a bit of heavy machinery that you probably don't know about yet.

To give you an idea though of what the definition will look like: it will say that everything in this world is a set hence $3$ is also a set. Hence when we define cardinality of a set to a set in bijection with the set. To give you an example, the cardinality of the natural numbers $\mathbb N$ is $\mathbb N$ itself. But the cardinality of the integers $\mathbb Z$ is also $\mathbb N$. And the cardinality of the rationals $\mathbb Q$ is also $\mathbb N$. In fact, every countable infinite set you can think of has cardinality $\mathbb N$.

I will stop here since going on into details will only be confusing at this stage.

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I never saw the axiom of choice used in the definition of cardinality. –  Asaf Karagila Dec 18 '12 at 21:06
I don't see your point, if you want all sets to have a cardinality then you need AC. Anyway, I'm not in the mood for a heated discussion. –  Matt N. Dec 18 '12 at 21:09
But cardinality is merely the equivalence relation defined by bijections. What you are saying means that there is absolutely no meaning to the phrase "The cardinality of the real numbers is $2^{\aleph_0}$" in ZF without choice, but alas there is a bijection between the real numbers and the power set of the natural numbers regardless to any assumptions of choice. –  Asaf Karagila Dec 18 '12 at 21:14
Happy now? If yes, can I have some peace and quiet? –  Matt N. Dec 18 '12 at 21:16
Yes, and yes. :-) –  Asaf Karagila Dec 18 '12 at 21:20

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