Questions tagged [cardinals]

This tag is for questions about cardinals and related topics such as cardinal arithmetics, regular cardinals and cofinality. Do not confuse with [large-cardinals] which is a technical concept about strong axioms of infinity.

Cardinality is a notion of size for sets, usually denoted by $|A|$ as the "cardinality of $A$". With finite sets the cardinality is simply the number of elements which are members of a set.

Dealing with infinite sets we can measure them in different ways. Cardinal numbers are very natural in the sense that they do not require extra structure (such as relations and operations defined on the set to be preserved).

In formal terms, suppose $f\colon A\to B$ (i.e. $f$ is a function whose domain is $A$ and its range is a subset of $B$).

We say that $f$ is injective if $f(a)=f(b)$ implies $a=b$; we say $f$ is surjective if its range is all $B$, namely for any $b\in B$ there is $a\in A$ such that $f(a)=b$.

If $f$ is both surjective and injective we say that $f$ is a bijection from $A$ to $B$. The inverse of a bijection is also a bijection.

Now we can define an equivalence relation on sets, $A\sim B$ if and only if there is some $f\colon A\to B$ which is a bijection.

Assuming the Axiom of Choice, we have that every set can be well ordered, and therefore there is a least ordinal which is equivalent to $A$, so we can assign it as a canonical representative for the equivalence class, usually denoted by $\aleph_\alpha$ where $\alpha$ is an ordinal, or as general Greek letters such as $\kappa,\lambda$.

Before defining the $\aleph$ numbers we need to define initial ordinals. Let $\alpha$ be an ordinal, if there is no $\beta<\alpha$ and $f\colon\alpha\to\beta$ which is a bijection, then $\alpha$ is called an initial ordinal.

The $\aleph$ numbers are defined by transfinite induction as:

  1. $\aleph_0 = |\omega| = \omega$ (note that $\omega$ is an initial ordinal),
  2. $\aleph_{\alpha+1} = \aleph_\alpha^+$ (where the $\cdot^+$ means the smallest initial ordinal above the one defined for $\aleph_\alpha$)
  3. If $\beta$ is a limit ordinal, then $\displaystyle\aleph_\beta = \bigcup_{\delta<\beta}\aleph_\delta$ (It is easy to verify that the union of initial ordinals is an initial ordinal).

The confinality of an $\aleph$ number is the minimal cardinality of a set which is unbounded in the initial ordinal matching the $\aleph$ number.

A cardinal is called regular if its cofinality is itself, otherwise it is called singular.

Example: $\aleph_0$ is regular, because for a set to be unbounded below $\omega$ it cannot be finite.

$\aleph_1$ is also regular, every ordinal below $\omega_1$ is countable, and the union of countably many countable ordinals is just countable - which is still below $\aleph_1$.

Example: $\aleph_\omega$ is singular, recall $\displaystyle\aleph_\omega=\bigcup_{n<\omega}\aleph_n$. Therefore the set $\{\omega_n\mid n<\omega\}$ (the collection of initial ordinals whose cardinality is less than $\aleph_\omega$) is unbounded, and its cardinality is merely countable.

The question whether or not there exists $\aleph_\delta$ such that $\delta$ is a limit ordinal, but $\aleph_\delta$ is regular is unprovable in ZFC. It is known that it is consistent that there are none, but unknown that it is inconsistent that there are. Cardinals with this property are called Large cardinals and are used for consistency proofs.


In the absence of choice we can no longer have canonical representatives for the equivalence classes, and things become tricky. The class of cardinals can still be defined, however in a slightly different way - usually Scott's trick.

However, to show how things can break down it is consistent with ZF that there is no choice function on the equivalence classes (i.e. you cannot have canonical representatives).

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Is there a set whose power set is countably infinite?

Does there exist a set whose power set is countably infinite? I know for sure that if a set has a finite number of elements, then its power set must have a finite number of elements, and if a set has an infinite number of elements, then its power…
user467365
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The cardinality of a countable union of countable sets, without the axiom of choice

One of my homework questions was to prove, from the axioms of ZF only, that a countable union of countable sets does not have cardinality $\aleph_2$. My solution shows that it does not have cardinality $\aleph_n$, where $n$ is any non-zero ordinal…
Zhen Lin
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What do the cosets of $\mathbb{R} / \mathbb{Q}$ look like?

$\newcommand{\R}{\Bbb R}\newcommand{\Q}{\Bbb Q}$ Looking at the group of real numbers under addition $(\R, +)$ it contains the (normal) subgroup of rational numbers $(\Q, +)$. I am wondering how to describe the cosets of $\R / \Q$. I know from…
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Are there number systems corresponding to higher cardinalities than the real numbers?

As most of you know, the set $\omega$ with cardinality $\aleph_0$ corresponds to what we normally know as the natural numbers $\mathbb{N}$, and the set $\mathcal{P}(\omega)$ with cardinality $\aleph_1$ corresponds to what we call the real numbers…
mrp
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Covering $\mathbb R^2$ with function graphs

Suppose we have a countable family of function graphs (each function is $\mathbb R\to\mathbb R$, not necessary continuous). Obviously, they cannot cover the whole plane $\mathbb R^2$, because they cannot even cover every of uncountably many points…
TauMu
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Why does the number of possible probability distributions have the cardinality of the continuum?

Wikipedia's article on parametric statistical models (https://en.wikipedia.org/wiki/Parametric_model) mentions that you could parameterize all probability distributions with a one-dimensional real parameter, since the set of all probability measures…
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Why hasn't GCH become a standard axiom of ZFC?

I've never seen a text that includes GCH in the ZFC axioms. I presume this means that GCH has not achieved widespread acceptance. This seems surprising to me, given that: The cardinal numbers encountered in "ordinary" mathematics are the beth…
goblin GONE
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There is no smallest infinity in calculus?

Somewhat of a basic question, but I tried mixing set theory and calculus and the result is a giant mess. From set theory (assume ZFC) we know there is a smallest infinite cardinal, $\aleph_0$, and that infinite numbers are well ordered, $\aleph_1 >…
Oria Gruber
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How to reach $\aleph_1$ without power sets?

At the end of Set Theory and the Continuum Hypothesis, Paul Cohen wrote: A point of view which the author feels may eventually come to be accepted is that CH is obviously false. The main reason one accepts the Axiom of Infinity is probably that we…
hmakholm left over Monica
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The Aleph numbers and infinity in calculus.

I have a fairly fundamental question. What is the difference between infinity as shown by the aleph numbers and the infinity we see in algebra and calculus? Are they interchangeable/transposable in any way? For example, could you describe the…
albertjq
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Why study cardinals, ordinals and the like?

Why is the study of infinite cardinals, ordinals and the like so prevalent in set theory and logic? What's so interesting about infinite cardinals beyond $\aleph _0 $ and $\mathfrak{c} $? It seems like they're enough for all practical purposes and…
user132181
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Does ZF prove that there are more cardinals than elements of any set?

In ZFC, it is easy to prove there are more cardinals than elements of any set. Specifically, given a set $X$, pick a well-ordering of $X$, and then you can inject $X$ into the cardinals by mapping its $\alpha$th element to $\aleph_\alpha$. My…
Eric Wofsey
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Medial Limit of Mokobodzki (case of Banach Limit)

A classical Banach limit is very useful concept for me, but there is a problem with the integration and even with the measurability, this means for a sequence $(f_n)_{n\in \mathbb{N}}$ of measurable (eg borel) and uniformly bounded functions, the…
Dawid C.
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Cardinality of a set A is strictly less than the cardinality of the power set of A

I am trying to prove the following statement but have trouble comprehending/going forward with some parts! Here is the statement: If $A$ is any set, then $|A|$ $<$ $|P(A)|$ Here is what I have so far: We need to show that there is an injection…
nicefella
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