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 real numbers

Given an open interval, say $(a,b]$, how do we show that it has the same cardinality as the set of real numbers? Or is there a bijective mapping in the 1st place?
Jus
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Cadinality of real sets using bijection

Prompt Let $a,b,c,d \in \mathbb R$ with $a < b $ and $c
Tsangares
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Properties of cardinals

How can I prove that $\operatorname{card}(n)=n$ for all $n \in \omega$, where we defined $$\operatorname{card}(x):= \min \{\alpha \in \operatorname{Ord} \mid \exists f: \alpha\to x\ \wedge f \text{ is bijection}\}$$ I am trying to use induction and…
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Rational numbers - countability

I have to show that the set of all finite sequences $$ (q_1,q_2,\dotsc,q_k),\quad k \in \mathbb{N} $$ of rational numbers is countable. To prove that the set $\mathbb{Q}$ of all rational numbers is countable, I used that the set…
Siri
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How many real-valued functions with one point of discontinuity are there?

I was given the following task: Find the cardinality of the set $C$ of real-valued functions with exactly one point of discontinuity. What I've found so far is that $$ \lvert \mathbb{R} \rvert \le \lvert C \rvert $$ because the map $f \colon…
user124708
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What alternative measures of infinite sets exist?

Cardinality measures the set of even numbers to be equal in size to the set of natural numbers, which I have no problem with. Are there accepted alternative measures of size which do recognise the fact that there are two natural numbers for every…
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Suppose $A$ and $B$ are sets with $\#A=\#\mathbb{Z}$ and $\#B=\#\mathbb{Z}$. (a) Prove $\#(A\cup B)=\#\mathbb{Z}$. (b) Is $\#(A\cap B)=\#\mathbb{Z}$?

Suppose $A$ and $B$ are sets with $\#A=\#\mathbb{Z}$ and $\#B=\#\mathbb{Z}$. Prove that $\#(A\cup B)=\#\mathbb{Z}$. Is it necessarily true that $\#(A\cap B)=\#\mathbb{Z}$? I'm sure that I need to use bijection to prove this, but I don't know the…
George S
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How to subtruct countable sequence from countable sequence

In a classic proof that the set of sequence points together with its limit is compact, we use the statement that any neighbourhood of a limit point contains infinite number of sequence points, hence we need only finite number of open sets to cover…
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Which of the following are countable?

Which of the following are countable? The set of all algebraic numbers The set of all strictly increasing infinite sequences of positive integers. The set of all infinite sequences of integers which are in arithmetic progression. My…
Learnmore
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$\mathcal P \left({\mathbb{R}}\right)$ has strictly greater cardinality than $\mathcal P \left({\mathbb{N}}\right)$

Show that $\mathcal P \left({\mathbb{R}}\right)$ has strictly greater cardinality than $\mathcal P \left({\mathbb{N}}\right)$ without using the arithmetic of infinite cardinals. This is a problem from Thomas Sibley's Foundations of Mathematics. It…
Lotte
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cardinality of cartesian product of the infinite set of natural numbers

One of the problems in my discrete math course states that we need to prove that $\mathcal{N}\times\mathcal{N}$ is countable specifically when there's a function $f:\mathcal{N}\times\mathcal{N}\to\mathcal{N}$ defined as…
Yos
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The usual bijection between $[0,1]$ and $[0,1]\times[0,1]$

While trying to explain to someone else how you can have a bijection between $[0,1]$ and $[0,1]\times[0,1]$, I found an issue in the usual bijection that we use. The usual bijection that I'm talking about is the one where $x=0.a_1a_2a_3a_4\ldots$ is…
H. Potter
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Cardinality of a union of uncountable and countable set.

Suppose you have a set A which has the same cardinality with real numbers R, which means |A| = |R|. Also suppose that you have a finite set B , which of course has finite cardinality. Also suppose A is a proper subset of A union B, I mean there are…
mehdi
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How to measure difference in size of infinite objects?

We know $\Bbb R$ is bigger than $\Bbb Q$ because its cardinality is bigger. We know that $\Bbb R^2$ is bigger than $\Bbb R$, which is bigger than $[0, 1]$ because the latter can be thought of as a subset of the former. However, the question is,…
Alex
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$A$ countable, $f :A \rightarrow B$ surjective. Prove $B$ is at most countable

My question is: can this be proven without using the almighty Axiom of Choice? Here's the idea of my proof using the axiom: We need an injective function from $B$ in $A$. Let $e$ be the choice function, $e: P(A) \rightarrow A$. It's easy to see that…