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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Validity of Proof for the Size of a Set Being Smaller Than Its Power Set

I am currently working on a proof regarding the cardinality of a set and its power set and would greatly appreciate some feedback and guidance. Let $A$ be a set and let $P(A)$ denote its power set. I have defined a function $f: A \rightarrow P(A)$…
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Compare cardinality of two sets

I had an argument about the following statement, which I think is wrong: One third of odd numbers (33.333%) will be divisible by 3. Let $\Bbb{N}$ refer to the positive integers. Let $\Bbb{O} := \{o | o = 2n-1; n \in \Bbb{N}\}$, the set of odd…
ste5an
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Let $F$ be the set of all finite subsets of $A$; prove that $\operatorname{card}(F)=\operatorname{card}(A)$

$A$ is an infinite set. Let $F$ be the set of all finite subsets of $A$, prove that $\operatorname{card}(F)=\operatorname{card}(A)$. I want to prove this conclusion: $\operatorname{card}\left(F\right)\leq \operatorname{card}\left(\bigsqcup_n…
Quyi Le
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Is $\aleph_0 - n$ defined for finite $n$?

The reason I am asking this is since $\aleph_0 - n$ = $- n+\aleph_0$, which would be undefined in cardinal arithmetic (since all cardinal numbers are positive). However, if we take two infinite sets $a$ and $b$, where $a$ has $n$ less elements than…
MathGeek
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Is it possible to construct a function $f: \mathbb{R} \rightarrow \mathbb{Z}$ such that $f$ is injective?

this was a true and false question which I mistakenly thought was false. My reasoning was thus: Let the function $f: \mathbb{R} \rightarrow \mathbb{Z}$ be injective. It follows that for every element $x \in \mathbb{R}$, there exists a unique $y \in…
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Showing $\lvert \mathbb{N}^\mathbb{R} \rvert = \mathcal{P}(\mathbb{R})$

What is the easiest way to show that $$ \lvert \mathbb{N}^\mathbb{R} \rvert = \mathcal{P}(\mathbb{R}) ? $$ I already know one cumbersome method, but I'd like to know if there is a simpler/cleverer way. If there's a technique that just uses basic…
Jeremy Lindsay
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Find the cardinality of the set $\{ m + \sqrt 5 n + \sqrt3 p : m, n, p \in\mathbb Q\}.$

I'm trying to find the cardinality of this set: $$\{ m + \sqrt 5 n + \sqrt3 p : m, n, p \in\mathbb Q\}.$$ However, this is somewhat a tougher set to calculate a cardinality for. How do we go about calculating the cardinality for this set?
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The set $\{x\in\mathbb R\mid x>0,(1+x^2)\tan2x=x\}$ is countably infinite

The set $\{x\in\mathbb R\mid x>0,(1+x^2)\tan2x=x\}$ is Empty Finite Countably infinite Uncountable I have plotted the graph and see that the required set is countably infinite. How to prove this fact?
vqw7Ad
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Given two sets $A$ and $B$, is it true that $|B^A|=\max\{|B|,|\mathcal{P}(A)|\}$?

My intuition is that this is true, because it makes sense for little sets (with cardinalities $|\mathbb{N}|,|2^\mathbb{N}|$). But i'm not sure at all that this is actually true. If it's not, what's the real relation between $|B^A|$ and…
R.V.N.
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Cardinality of $ \mathbb R \times \mathbb R $ and $ \mathbb R $.

My attempt is based on giving an explicit function between both sets but through compositions, which is up to what I have learned so far. First, let us consider the map $ f: (0,1) \times (0,1) \to (0,1) $, defined for two real numbers $ x, y \in…
JamieG
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The meaning of GCH in ZF

I heard one can prove that GCH implies AC(the axiom of choice) in ZF. But I am confused with the meaning of GCH in ZF. Under AC, we can define cardinal exponentiation, so the cardinal 2 to the aleph is well defined. However, Without AC, How can we…
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Trouble understanding cardinality

Hi guys I am having trouble understanding cardinality. I am given this practice question. 1) Use Cantor-Schroder-Bernstein Theorem to prove that the intervals $(0,1)$ and $[0,1]$ have the same cardinality? My attempt : I know the CSB theorem is Let…
MathGeek
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$X = \{(a_n): a_n \in \mathbb{N} \text{ and } a_n \leq a_{n+1}$ for every $n \in \mathbb{N}\}$

Let $X = \{(a_n): a_n \in \mathbb{N} \text{ and } a_n \leq a_{n+1}$ for every $n \in \mathbb{N}\}$ Prove $$\vert X\vert = \vert \mathbb R\vert=c$$ without using $2^{\aleph_0} = \aleph_0^{\aleph_0} = c$ My Attempt I know that $\vert X\vert = \vert…
Silkking
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Prove that an interval of rationals has the same cardinality as the rationals

I am trying to find an answer to this problem "Show that the sets $\mathbb Q$ of rational numbers and $A = \{x \in\mathbb Q | 0 \le x \le 1\}$ have the same cardinality" So, I know we would have to find a bijection from $\mathbb Q$ to $A$. I was…
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Proving that the cardinality of the set of all bijective functions from $\mathbb N$ to $\mathbb N$ is $c$

I was curious about this one the other day. I came up with my own proof but it's reliant on the continuum hypothesis. The proof is below, but I was wondering if it's possible to have a "proven" proof. Also, I want to get the proof below reviewed to…
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