Once we construct the definition of the ordinals:

$$0=\{\} \, \, 1=\{0\} \,\, 2=\{0,1\} \,\,3= \{0, 1,2\} \,\, ...$$

And we want to describe the cardinality of the set $S$:


Intuitively we know that:


However, how do we describe the function of $\mathbb{card}(x)$? This function must map the Von-Neumann ordinals to the cardinality of the set. How does this function do that? If the cardinality of the set can be described by an ordinal, why do we denote:

$$|\mathbb{N}| = \aleph_0$$

And not:

$$|\mathbb{N}| = \omega_0$$

What would be the ordinal number that would be associated with $\aleph_1$. I thought that it could be $\omega_0 + 1$ since it is the next size of infinity. Like the next size in cardinal arithmetic is always just adding one. Or the idea of the next. However, if it is not that what would it be? Would the construction of the cardinality function help, or not?

If this is not clear, please let me know and I'll edit this a bit more.


  • $\begingroup$ It depends how you contruct the numbers. If you start from set theory, then the number 0 is defined as the cardinality of $\emptyset$, then number 1 is defined as cardinality of $\{ \emptyset \}$ and go on . $\endgroup$
    – L F
    Aug 31, 2020 at 19:02
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    $\begingroup$ The size of the reals isn't necessarily the one that comes after the naturals. We can't construct any cardinalities that are probably in-between the two, but that doesn't mean there aren't any. $\endgroup$
    – Arthur
    Aug 31, 2020 at 19:03
  • $\begingroup$ We have several definitions here: ordinals, natural numbers, real numbers, cardinals. You can't make more definitions with those. Now it's time to prove claims such as "this will also be that". $\endgroup$
    – Asaf Karagila
    Aug 31, 2020 at 19:03
  • $\begingroup$ @LuisFelipe How is the function that maps the number of elements of the set to the natural numbers defined? $\endgroup$ Aug 31, 2020 at 19:07
  • $\begingroup$ @Arthur I've updated my question, to address the issue that you pointed out. What I'm really asking, is it possible to define cardinals with ordinal numbers. $\endgroup$ Aug 31, 2020 at 19:08

1 Answer 1


As the Wikipedia page notes, there are two ways to approach cardinality. One, which is what you are getting at here, is to construct the cardinal numbers and have a procedure that assigns each set $S$ a unique cardinal $Card(S)$. This construction is somewhat involved and is mostly the domain of set theorists and the like. Most ordinary mathematicians think of cardinality through the relations "$A$ has the same cardinality as $B$" denoted by $|A| = |B|$ and "$A$ has cardinality less than or equal to $B$", denoted by $|A| \leq |B|$.

We define $|A| = |B|$ as $\exists \phi:A \to B, \phi$ is a bijection. And we define $|A| \leq |B|$ by $\exists \phi: A \to B, \phi$ is an injection.

Then the Schroder-Bernstein theorem gives that $|A| \leq |B|$ and $|B| \leq |A| \implies |A| = |B|$.

Now if we just consider finite sets, we can alternatively define a "function" (note it won't be a true set function, since there is no set of all finite sets) $Card(S)$ that assigns a finite set $S$ a unique natural number that is its number of elements. We then can note that $Card(A) = Card(B) \iff |A| = |B|$ and $Card(A) \leq Card(B) \iff |A| \leq |B|$, so these two approaches are the same for finite sets.

Edit: Defining the $Card$ "function" for finite sets. Since $Card$ cannot be a set function as noted above, we are really looking for a predicate $\Phi(A,n)$ s.t. $A$ is finite implies $\exists! n \in \mathbb{N}, \Phi(A,n)$.

Denote $set(n) = \{0,...,n-1\}$. Define $\Phi(A,n) \iff n \in \mathbb{N} \land \exists \phi : A \to set(n), \phi$ is a bijection.

Then to show $\Phi$ has the properties we want, we note that uniqueness comes directly from the nonexistence of bijections between $set(n)$ and $set(m)$ for $n \neq m$ and since $A$ is finite can be defined to mean $\exists n \in \mathbb{N} \exists \phi : A \to set(n), \phi$ is a bijection, we get existence of some $n$ s.t. $\Phi(A,n)$ provided $A$ is finite.

Thus $\Phi$ defines a function, since to each finite $A$, we get a unique $n \in \mathbb{N}$.

  • $\begingroup$ How is the function constructed? Is there a formula for it? $\endgroup$ Aug 31, 2020 at 19:17
  • $\begingroup$ I'll make an edit explaining that in some detail. $\endgroup$ Aug 31, 2020 at 19:19
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    $\begingroup$ @KeeferRowan Ok, thank you!!! $\endgroup$ Aug 31, 2020 at 19:47
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    $\begingroup$ The ordinals and cardinals are constructed differently to serve different purposes. The ordinals numbers are supposed be a canonical form for order type of well ordered sets (i.e. every well ordered set is has an order isomorphism to one and only one ordinal) while cardinals are supposed be a canonical form for cardinality (i.e. every set has a bijection to one and only one cardinal). There constructions reflect this aim. Clearly every ordinal has a cardinality, but different ordinals have the same cardinality, e.g. $|\omega| = |\omega +1|$. $\endgroup$ Aug 31, 2020 at 19:56
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    $\begingroup$ On the other hand, it only makes sense to assign an ordinal to a well-ordered set, and cardinals don't have an a priori order associated with them. That they can be well-ordered is a consequence of the axiom of choice (more specifically, the well-ordering principle). Once you choose a well ordering, there will be some ordinal that corresponds to a given cardinal, but this won't be unique (since for any infinite cardinality, multiple ordinals will represent it, since for any infinite ordinal $\nu$, $|\nu| =|\nu +1|$). $\endgroup$ Aug 31, 2020 at 19:59

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