Overview of basic results on cardinal arithmetic

Question

Are there some good overviews of basic formulas about addition, multiplication and exponentiation of cardinals (preferably available online)?

Answer 1

$\newcommand{\alnul}{\aleph_0}\newcommand{\mfr}[1]{\mathfrak{#1}}\newcommand{\Ra}{\Rightarrow}\newcommand{\card}[1]{\left|#1\right|}\newcommand{\powerset}[1]{\mathcal P(#1)}\newcommand{\Lra}{\Leftrightarrow}\newcommand{\Zobr}[3]{#1\colon#2\to#3}$I have no doubt that you there are many useful online resources for these, but many such identities are available here at MSE, together with their proofs.

I'll give a list of some basic results on cardinal arithmetics and I'll add links to results, which have proofs here at MSE. I am making this CW, so feel free to add more identities and pointers to further useful questions and answers.

In the identities bellow, $a$, $b$, $c$ denote arbitrary cardinals, $X$ is an arbitrary set, $\alnul$ is the cardinality of $\mathbb N$ and $\mfr c=2^{\alnul}$. Cardinality of a set $X$ is denoted by $\card X$ and $\powerset X$ is the notation of the power set of $X$.

Equality of cardinal numbers is defined as follows: $$|A|=|B| \Lra \text{ there exists a bijection }\Zobr fAB.$$ Inequality of cardinal numbers is defined as follows: $$|A|\le|B| \Lra \text{ there exists an injective function }\Zobr fAB.$$ The definitions of the operations on cardinal numbers (addition, multiplication, exponentiation) can be found e.g. in this answer.

Validity of Axiom of Choice is assumed. If you want to learn about cardinals without AC, you can have a look e.g. at this question: Defining cardinality in the absence of choice

• $a\le b$ $\land$ $b\le c$ $\Ra$ $a\le c$

This follows from the fact that composition of two injective maps is injective

• $a\le b \land b\le a \Ra a=b$

This is just a reformulation of Cantor-Bernstein's theorem.

• For any two sets $A$, $B$ we have $|A|\le|B|$ $\lor$ $|B|\le|A|$. I.e. any two sets/any two cardinal numbers are comparable.

Note: Proof of this result uses the Axiom of Choice. For the role of AC in this result see here: Is the class of cardinals totally ordered? and For any two sets $A,B$ , $|A|\leq|B|$ or $|B|\leq|A|$.

• $|A|\le|B| \Lra (\text{there exists a surjective function }\Zobr fBA \text{ or $A$ is empty}).$

There exists an injection from $X$ to $Y$ if and only if there exists a surjection from $Y$ to $X$.

Note: Proof of this result uses the Axiom of Choice.

• $\card{\powerset X}=2^{\card X}$

See e.g. How to show equinumerosity of the powerset of $A$ and the set of functions from $A$ to $\{0,1\}$ without cardinal arithmetic? or this answer. The question What is the set of all functions from $\{0, 1\}$ to $\mathbb{N}$ equinumerous to? deals with a special case, but it can be easily generalized.

• $a+b=b+a$

This follows simply from commutativity of union: $A\cup B=B\cup A$.

• $a+(b+c)=(a+b)+c$

This follows from associativity of union: $A\cup(B\cup C)=(A\cup B)\cup C$.

• $b\le c \Ra a+b\le a+c$

This is (after adding some details) basically the same thing as the implication $B\subseteq C$ $\Ra$ $A\cup B\subseteq A\cup C$.

• $ab=ba$

A bijection $A\times B\to B\times A$ can be given by $(x,y)\mapsto (y,x)$.

• $a(bc)=(ab)c$

A bijection between $A\times(B\times C)$ and $(A\times B)\times C$ can be given by $(x,(y,z))\mapsto ((x,y),z)$.

• $a(b+c)=ab+ac$

This follows from the fact that $A\times(B\cup C)=A\times B\cup A\times C$.

• $b\le c \Ra ab\le ac$

See e.g. Proof of cardinality inequality

• $a^2=a\cdot a$

See e.g. this answer

• $a\le b \Ra a^c\le b^c$

See e.g. this answer.

• $a\le b \land c\ne 0 \Ra c^a\le c^b$

Note that this is not true for $c=0$, since $0^0=1$. (The set $\emptyset^\emptyset=\{\emptyset\}$ has one element.) The set $\emptyset^\emptyset$ and its cardinality is also discussed here.

• $a^{b+c}=a^b\cdot a^c$

See e.g. Notation on proving injectivity of a function $f:A^{B\;\cup\; C}\to A^B\times A^C$.

• $(a^b)^c=a^{bc}$

See e.g. How to show $(a^b)^c=a^{bc}$ for arbitrary cardinal numbers?

• $a^b\le 2^{ab}$

See e.g. this answer

• $a<2^a$

This is Cantor's theorem. The question Is the class of subsets of integers countably infinite? deals with the special case $a=\alnul$, but there are answers which discuss the more general result or can be easily generalized.

• $\alnul+\alnul=\alnul$

See e.g. Let $X$ and $Y$ be countable sets. Then $X\cup Y$ is countable

• $a\ge\alnul \Ra \alnul+a=a$

• $\alnul\cdot\alnul=\alnul$

See e.g. Bijecting a countably infinite set $S$ and its cartesian product $S \times S$, How does one get the formula for this bijection from $\mathbb{N}\times\mathbb{N}$ onto $\mathbb{N}$?, The cartesian product $\mathbb{N} \times \mathbb{N}$ is countable and Proving the Cantor Pairing Function Bijective

• $\alnul^{\alnul}=2^{\alnul}=\mfr c$

See e.g. Is $\aleph_0^{\aleph_0}$ smaller than or equal to $2^{\aleph_0}$?

• If $a$ is infinite cardinal, then $a^2=a$.

See e.g. About a paper of Zermelo

Note: Proof of this result uses the Axiom of Choice.

• If $b$, $c$ are infinite cardinals, then $b+c=bc=\max\{b,c\}$.

See e.g. Simple cardinal arithmetic

Note: This is a consequence of the preceding result, so it relies on the Axiom of Choice too.