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 equivalence classes

The relation ~ is defined on P(N): A~B if |A| = |B|. I need to prove that the cardinality of the equivalence classes is countable. Any ideas??
Itay4
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Infinite cardinals formal proof

I need help in showing $$2^a + 2^a = 2 · 2^ a$$ It sounds obvious but how can I formally prove it?
user401516
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Set of all functions of $\mathbb N$ $\to$ $\mathbb R$

I got a question about the cardinal of set of all function of $\mathbb N$ $\to$ $\mathbb R$. I think that the cardinal number of the set is C because the range of all function is R which the cardinal number of it is C, but I dont knowhow to prove…
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SCH,singular cardinal hypothesis, GCH, relationship

Here it is stated that SCH is a consequence of GCH and also SCH is whether GCH might fail at a singular cardinal. These two condition seem yield a contradiction:does SCH follow from failing or holding GCH at singular cardinals?
user122424
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Proving that a particular alphabet is a denumerable set

Suppose that we have a written language consisting of all 26 letters of the English alphabet. Words are formed in this language by constructing finite strings of integers. a) Prove that the set of all words W, in this language is a denumerable…
Owen
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What is $n^{\aleph_0},n\in\mathbb N$

Can I say that $$n^{\aleph_0}=2^{\aleph_0\log_2n}=2^{\aleph_0}=\aleph_1$$
RE60K
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How to create a bijection from the union of the set natural numbers and square root 2 to the set of natural numbers?

Since $\Bbb N$ and ${\sqrt2}$ are each countable sets, I see that the union is also countable. From this and the fact that $\Bbb N$ union ${\sqrt2}$ is infinite we know there exists a bijection to $\Bbb N$ I understand why such a bijection exists,…
user319635
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Are the numbers $\beth_n$ for $n > 0$ signed or unsigned?

By extending the real number line in both directions, I know that $\infty$ or $\aleph_0$ or whatever else you want to call it has a negative, i.e. $-\aleph_0$ is a thing. Now, of course, $\beth_0 = \aleph_0$, so $-\beth_0$ is also a thing in this…
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Prove that $\{X \in P(Z)| X \text{ is finite}\}$ is enumerable.

I am not sure how to approach this problem. if you could help it would be great.
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Prove that $(0,1) =c\mathbb R$

How do I prove this knowing that $f(x) = \tan(x\pi/2)$ is a bijection between $(0,1)$ and $(0, \infty)$? We also have a bijection between $(-1,1)$ and $(0,1)$.
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Confused by how to proof some statements about cardinals

I have a set of statements such as: Proof $\aleph_0+\aleph_0=\aleph_0$ I know that $|\Bbb Z|=\aleph_0$ and that for countable $A,B$ $A\cap B=\emptyset$: $|A\cup B|=|A|+|B|$. To this I add that if $A=\{-1,-2,...\}$ and $B=\Bbb N$,…
YoTengoUnLCD
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what is the cardinality of powerset of a union set?

Is there exist something like P(X+Y) (P STANDS FOR POWERSET)? I am confuse because power set is the set of all subset of Cartesian product, and X+Y wont give Cartesian product but (x,0) U (y,1), and if it exist what is the cardinality of that power…
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smallest cardinal greater than an infinite ordinal is a regular cardinal

let $\alpha$ be an infinite ordinal, and $\alpha^+$ be the smallest cardinal greater than $\alpha$. Show that $\kappa^+$ is a regular cardinal. This is for homework, but I'm not really sure where to begin.
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Showing that $\aleph_0^{|2^\mathbb{N}|} > |2^\mathbb{N}|$

I attempted to prove that there's a surjective function $\mathbb{N} ^{|2^\mathbb{N}|} \rightarrow 2^\mathbb{N}$ and that there isn't one in the opposite way, however I wasn't able to prove the former. Is there another way to prove this cardinal…
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