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 $+\infty$?
Is it viable to ask in an infinite set about the Cardinality?
-
5Philosophically 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. – 2012-12-16
2 Answers
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 $\mathcal P(\mathbb N)=\{A\mid A\subseteq\mathbb N\}$.
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.
-
0Just to set a fact straight, for the record: one can define cardinality perfectly well without choice. That is, one can define in ZF a canonical class acting as a quotient of the equivalence relation defined by bijection. One way is known as “[Dana] Scott’s trick”: define $\|A\| := \{ X \in V_\alpha\ |\ X \cong A\}$, where $\alpha$ is the minimal ordinal such that *some* set in $V_\alpha$ has some bijection with $A$. (Toposes, algebraic set theory, whether or not we should care about uses of choice, and so on are all fun issues, but in this particular case, they’re beside the point.) – 2014-01-09