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).

3591 questions
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Cardinality of the Set of all finite subset of $\mathbb{R}$

Find the cardinality of the set of all finite subsets of $\mathbb{R}$. I have proved that the set of all finite subsets of $\mathbb{N}$ is countable . But I cannot find the cardinality of the set in case of $\mathbb{R}$ . My Attempt: First I have…
sigma
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A simple bijection between $\mathbb{R}$ and $\mathbb{R}^4$ or $\mathbb{R}^n$?

How to form a bijection from $(0,1]$ to $\mathbb{R}$: $$f(x) = \left\{\begin{array}{ll} 2-\frac{1}{x}&\text{if }x\in(0, .5]\\ \frac{2x-1}{1-x}&\text{if }x\in(.5, 1]. \end{array}\right.$$ So, to go from $\mathbb{R}$ to $\mathbb{R}^4$ shouldn't be…
futurebird
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Cardinal number subtraction

We all know that $|\mathbb{N}| = \aleph_0$. Since $|\{-1\} \cup \mathbb{N}| = \aleph_0$ as well, I guess you could say that $\aleph_0 + 1 = \aleph_0$. You can go on to derive that $\aleph_0 + \aleph_0 = \aleph_0$, and by induction $n \times \aleph_0…
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Cardinality != Density?

I was in a discussion where I argued that the density of two sets of the same cardinality could be different in respect to the infinite range of non-negative integers. Does cardinality imply that any set of ${\aleph_0}$ has equal density to any…
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Cardinal arithmetic gone wrong?

I am trying to calculate $\kappa^\lambda = \aleph_{\omega_1}^{\aleph_0}$. I know that if $\kappa$ is a limit cardinal and $0 < \lambda < \mathrm{cf}(\kappa)$ then $\kappa^{\lambda} = \displaystyle \sum_{\alpha < \kappa} |\alpha|^{\lambda}$. Hence…
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Why is ${\aleph_\omega}^{\aleph_1} = {\aleph_\omega}^{\aleph_0} \cdot {2}^{\aleph_1}$?

I am supposed to prove that ${\aleph_\omega}^{\aleph_1} = {\aleph_\omega}^{\aleph_0} \cdot {2}^{\aleph_1}$ , but I really have no idea how to start or what to do. I thought I could use the following fact: ${2}^{\aleph_1}= {\aleph_1}^{\aleph_1}$,…
Suze
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Cardinality of the set of permutations of a set $ A $

I've some trouble calculating the cardinality of the set of the permutations of a given set $ A $. For notational purpose let $ k = |A|$ and define $ P_A = \{ f : A \to A | f \text{ is a bijective function } \} $ Clearly if $ A $ is a finite set,…
Riccardo
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If $\aleph_\lambda>2^{\aleph_0}$ is a limit cardinal, then $\aleph_\lambda^{\aleph_0}=\aleph_\lambda$?

I know that for successor cardinals, the result holds; i.e. if $\aleph_\alpha$ is a successor cardinal, then because this implies that it is a regular cardinal, then we can use this to show that if $\aleph_\alpha\geq 2^{\aleph_0}$, then…
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Bijection between $\mathbb R^\mathbb N$ and $\mathbb R$

$\mathbb R^\mathbb N$ is the set of all functions from the naturals to the reals. I have to prove that $\mathbb R^\mathbb N$ has the same cardinality as $\mathbb R$. I found an injective function from $\mathbb R$ to $\mathbb R^\mathbb N$, but can't…
Ryan
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Prove that the intervals $(0,1)$ and $(0,\infty)$ have the same cardinality

I have to proof that the intervals $(0,1)$ and $(0,\infty)$ have the same cardinality. I find some similar example with $(0,1)$ and $\mathbb{R}$ but I still have no idea to solve it.
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Cardinality of a set of non-continuous functions

I have to find the cardinality of the set of the non-continuous functions $f:\mathbb{R}\rightarrow\mathbb{R}$. I think we should look for function that at least have a point of discontinuity, but i don't have a clue, really. Any hint?
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Why $|\Bbb{R}|=2^{\aleph_0}$ and not $|\Bbb{R}|=10^{\aleph_0}$?

Suppose we have a real number of this form: $$...x_3x_2x_1,x_{-1}x_{-2}x_{-3}...$$ Since $x_i\in\{0,1,2,3,4,5,6,7,8,9\} \ \ \ \forall i$. We have 10 choices for every number, this means that: $$|\Bbb{R}|=10^{|\Bbb{Z}|}$$ But since…
Kandinskij
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Prove that there is no injection from $\Bbb{R}$ to $\Bbb{N}$.

Knowing that there is no surjection from $\Bbb{N}$ to $\Bbb{R}$, I also know that there is no injection going the other way. I know how to prove that there is no injection from the power set of $S$ to $S$, but how can I prove that there is no…
Jim H
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$f : [0, 1] \rightarrow \text{ cardinals }$

Collection of all cardinality can not be a set. Can we find $f : [0, 1] \rightarrow \text{ cardinals }$ s.t. $f$ is 1-1 ?
Mudream
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Proof of Operation of Cardinals a,b : $(ab)^c = a^cb^c$

$\forall set\; A,B,C\;\;$Let $a=card(A),\;\; b=card(B),\;\; c=card(C)$ Claim prove $(ab)^c = a^cb^c$ We need to show that $(A\times B)^C \approx A^c \times B^c$. Thus let $\pi_A:A\times B\rightarrow A\;$ $\pi_B: A\times B \rightarrow B$ be the…
Daschin
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