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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Cardinality of a set of functions from $\mathbb N$ to a set of real numbers

Let $S$ be the set of functions from $\mathbb N$ to the set $\{\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{7}\}$. Determine $|S|$. I understand how to prove the converse case where $T$ is the set of functions from $\{\sqrt{2}, \sqrt{3}, \sqrt{5},…
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(S is a proper class wrt model M) implies ((|T|=|S| under model N) implies (T is a proper class wrt M)). Why?

If I understand correctly (which is far from guaranteed, so a reply telling me that it is rubbish will be better than no reply at all): Suppose we have two models N and M such that N is a (non-conservative) extension of M, such that the class S is…
nomadreid
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Cardinality of the set of differentiable functions

Is the cardinality of $$X = \{f: \Bbb R \to \Bbb R \;|\; f \text{ is differentiable everywhere}\}$$ the same as $\Bbb R$? How to prove it?
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Bijection between $ A ^ {B \cup C} $ and $ A ^ B \times A ^ C $, with $B\cap C=\varnothing$.

Let us consider $ f | _B $ the function restricted to $ B $ and analogously $ f | _C $. The formula for both is $ f | _B (b) = f (b) $ and $ f | _C (c) = f (c) $, for all $ b, c \ in B, C $ respectively. Let's define $ F: A ^ {B \cup C} \to A ^ B…
James A.
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Under what circumstances does $\kappa^\nu = 2^\nu$?

This question seems to show up a lot. Given cardinal numbers $\kappa$ and $\nu$, under what circumstances do we have $\kappa^\nu = 2^\nu$? Assume $\nu$ is infinite.
goblin GONE
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cardinality of functions from N to N using Schröder–Bernstein theorem

Im trying to prove that $\left|A\right|=\aleph$ for the following group: $A=\left\{f\in\mathbb{N}\rightarrow\left.\mathbb{N}\right|\forall n\le m\ .f\left(n\right)\le f\left(m\right)\right\}$ using Schröder–Bernstein theorem To prove that…
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(Proof) Order of Cardinals : $a\le c$, $b\le d$ $\rightarrow$ $a+b\le c+d$

claim $\forall \;set \; A,B,C,D$ Let $a =card(a), b=card(b), c=card(C), d=card(D)$ then $a\le c, b\le d$ $\Rightarrow$ $a+b\le c+d$ Proof $\exists \text{ injection } f: A \rightarrow C $ since $a\le c$ and similarly, $\exists \text{ injection } g: B…
Daschin
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Finding a singluar cardinality

Given an infinite cardinal b such that $b^2=b$ I need to find a cardinal $x$ such that: $$(x^b = 2^b)\quad \land \quad (\forall c.(c>x)\to (c^b>2^b))$$ Furthermore, I need to show there is only one such $x$. I think the answer is $ x = 2^b$ but…
user401516
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Summation over uncountable set while proving that cardinality of functions is bigger than $\Bbb{R}$

I'm currently in my first year of studying mathematics. In one of the subjects we learnt about the cardinality of different sets, among which $\Bbb{N}$ and $\Bbb{R}$. We also discussed the Power Set $P(A)$ as the set of all subsets of $A$ and proved…
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Ordering power sets by cardinality

Given two sets/classes $X,Y$ such that $\wp X\simeq \wp Y$ under bijection $g$ say, is there a bijection $h:\wp X \to \wp Y$ such that $|h(A)|=|A|$ for every $A\subseteq X$? Does the assumption $X\prec Y$ help with the proof? I think the statement…
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Cardinality of $A^B$ when $A > B \ge \aleph_0$

For an infinite cardinal A, then if $B$ is finite $A^B = A$ If $B$ is infinite and $B \ge A$ then $A^B \ge A^A \ge 2^A > A$ What if $B$ is infinite, but $B < A$, i.e. $A > B \ge \aleph_0$. Is there a relationship between $A^B$ and $A$ in this case…
Tom Collinge
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Cardinality of $P_\mathfrak{c}(\mathbb R)$

Calculate the cardinality of $\mathcal P_\mathfrak{c}(\mathbb R)=\{\mathcal R \in \mathcal P(\mathbb R) \, /\, \#(\mathcal R)=\mathfrak{c}\}$. I thought that instead of $\mathcal P_\mathfrak{c}(\mathbb R)$ I should think in $\mathcal P_c([0,1))$ so…
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What mean $L(\mathbb{R})$ and $L(\mathbb{R})^*$?

I found them relating a cardinality question here. Does it have anything to do with regularity/computability?
peterh
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Prove that for any infinite set $A$, $|\mathbb{N}|\le |A|$

How can you show that for any infinite set $A$, $|\mathbb{N}|\le |A|$? thanks
maxuel
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$\mu^\lambda = 2^\lambda$ for infinite $\mu$ and $\lambda$

The original question is (from Mathematical Logic, A course with Exercises): Assume Axiom of Choice, let $a$ and $b$ be 2 infinite sets with $\text{card}(a)=\lambda$ and $\text{card}(b)=\mu$. Assume $\lambda > \mu$. Let $g$ be an injective mapping…
Y.X.
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