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 cardinality a well defined function?

I was wondering if the cardinality of a set is a well defined function, more specifically, does it have a well defined domain and range? One would say you could assign a number to every finite set, and a cardinality for an infinite set. So the range…
Dirkboss
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How many compact Hausdorff spaces are there of a given cardinality?

This is a question I found myself wondering about recently. I eventually figured out the answer myself, but as this doesn't seem to be written down anywhere easy to find on the Internet I decided to share it here. Let $\kappa$ be an uncountable…
Eric Wofsey
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Cardinality of all cardinalities

Let $C = \{0, 1, 2, \ldots, \aleph_0, \aleph_1, \aleph_2, \ldots\}$. What is $\left|C\right|$? Or is it even well-defined?
kennytm
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Cardinality of a Hamel basis of $\ell_1(\mathbb{R})$

What is the cardinality of a Hamel basis of $\ell_1(\mathbb R)$? Is it deducible in ZFC that it is seemingly continuum? Does it follow from this that each Banach space of density $\leqslant 2^{\aleph_0}$ has a Hamel basis of cardinality continuum…
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What is the largest set for which its set of self bijections is countable?

I recently came across a problem which required some knowledge about the self bijections of $\mathbb{N}$, and after looking up how to construct some different bijections I came across the result that the set of self bijections of $\mathbb{N}$ is…
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The cartesian product $\mathbb{N} \times \mathbb{N}$ is countable

I'm examining a proof I have read that claims to show that the Cartesian product $\mathbb{N} \times \mathbb{N}$ is countable, and as part of this proof, I am looking to show that the given map is surjective (indeed bijective), but I'm afraid that I…
Harry Williams
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Defining cardinality in the absence of choice

Under ZFC we can define cardinality $|A|$ for any set $A$ as $$ |A|=\min\{\alpha\in \operatorname{Ord}: \exists\text{ bijection } A \to \alpha\}. $$ This is because the axiom of choice allows any set to be well-ordered, so that the set after…
LostInMath
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cardinality of all real sequences

I was wondering what the cardinality of the set of all real sequences is. A random search through this site says that it is equal to the cardinality of the real numbers. This is very surprising to me, since the cardinality of all rational sequences…
Vishal Gupta
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Do there Exist Proper Classes that aren't "Too Big"

Some proper classes are "too big" to be a set in the sense that they have a subclass that can be put in bijection with $\alpha$ for every cardinal $\alpha$. It is implied in this post that every proper class is "too big" to be a set in this sense,…
Stella Biderman
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If the infinite cardinals aleph-null, aleph-two, etc. continue indefinitely, is there any meaning in the idea of aleph-aleph-null?

If the infinite cardinals aleph-null, aleph-two, etc. continue indefinitely, is there any meaning in the idea of aleph-aleph-null? Apologies if this isn't a sensible question, I really don't know too much about these infinite cardinals aside from…
aleph_aleph_null
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How to formulate continuum hypothesis without the axiom of choice?

Please correct me if I'm wrong but here is what I understand from the theory of cardinal numbers : 1) The definition of $\aleph_1$ makes sense even without choice as $\aleph_1$ is an ordinal number (whose construction doesnt depend on the axiom of…
M.G
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Do you need the Axiom of Choice to accept Cantor's Diagonal Proof?

Math people: It is my understanding that set theorists/logicians compare cardinalities of sets using injections rather than surjections. Wikipedia defines countable sets in terms of injections. Cantor's diagonal proof that the real numbers are…
Stefan Smith
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Proof that the number of proofs is countably infinite

I think that the number of mathematical statements is countably infinite, since each statement is a finite string of finitely many symbols, but i am not sure how to prove it. Once i prove that, i can say that a every math proof with $n$ steps is a…
yes
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How to show $(a^b)^c=a^{bc}$ for arbitrary cardinal numbers?

One of the basic (and frequently used) properties of cardinal exponentiation is that $(a^b)^c=a^{bc}$. What is the proof of this fact? As Arturo pointed out in his comment, in computer science this is called currying.
Martin Sleziak
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For any two sets $A,B$ , $|A|\leq|B|$ or $|B|\leq|A|$

Let $A,B$ be any two sets. I really think that the statement $|A|\leq|B|$ or $|B|\leq|A|$ is true. Formally: $$\forall A\forall B[\,|A|\leq|B| \lor\ |B|\leq|A|\,]$$ If this statement is true, what is the proof ?
Amr
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