# Overview of basic results on cardinal arithmetic

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


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 $|B|\leq|A|$ and Proving $(A\le B)\vee (B\le A)$ for sets $A$ and $B$.

• $|A|\le|B| \Lra (\text{there exists a surjective function }\Zobr fBA \text{ or$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 Finding a correspondence between $\{0,1\}^A$ and $\mathcal P(A)$ 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 $B\subseteq C$ $\Ra$ $A\cup B\subseteq A\cup C$. See Does $a \le b$ imply $a+c\le b+c$ for cardinal numbers?

• $ab=ba$

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 $(x,(y,z))\mapsto ((x,y),z)$. See here: A bijection between $X \times (Y \times Z)$ and $(X \times Y) \times Z$

• $a(b+c)=ab+ac$

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

• $b\le c \Ra ab\le ac$
• $a^2=a\cdot a$

• $a\le b \Ra a^c\le b^c$

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

See e.g. this question.

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.

• $a^{b+c}=a^b\cdot a^c$
• $(a^b)^c=a^{bc}$
• $(ab)^c=a^c\cdot b^c$
• $a^b\le 2^{ab}$

• $a<2^a$

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. See also Understanding the proof for $2^{\aleph_0} > \aleph_0$. This question asks about the general result: Cardinality of a set A is strictly less than the cardinality of the power set of A

• $\alnul+\alnul=\alnul$
• $a\ge\alnul \Ra \alnul+a=a$

• $\alnul\cdot\alnul=\alnul$

The following result is, to some extent, related to the following:

• Union of countably many countable sets is countable.

It is worth mentioning that proof of this uses Axiom of Choice. See Prove that the union of countably many countable sets is countable.

• $\alnul^{\alnul}=2^{\alnul}=\mfr c$

See e.g. Is $\aleph_0^{\aleph_0}$ smaller than or equal to $2^{\aleph_0}$? or What is $\aleph_0$ powered to $\aleph_0$? or What's the cardinality of all sequences with coefficients in an infinite set? (One of the answers to this question also discusses powers of the form $\aleph_\alpha^{\aleph_\beta}$ in general.

• 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. See also: For every infinite $S$, $|S|=|S\times S|$ implies the Axiom of choice

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

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

• I've posted a question at meta about the format of answer I've chosen here. meta.math.stackexchange.com/questions/3967/… – Martin Sleziak Apr 13 '12 at 7:31
• Do you think that you should mention the use of the axiom of choice, or at least which results holds without it? – Asaf Karagila Apr 13 '12 at 8:12
• @Asaf: I've added a note mentioning AC. (The last two are the only identities from the list that use AC, right?) To be honest, I am not completely sure about the usual definition of cardinal and cardinality when we do not assume AC. I should probably read one of your answers on this topic. ;-) – Martin Sleziak Apr 13 '12 at 8:20
• BTW I was a little surprised that I did not find the proof of $a^2=a$ here. (Maybe we don't have it here, maybe it is somewhere but I did not find it.) – Martin Sleziak Apr 13 '12 at 8:24
• You have set the bar high for future "Overview" questions! – The Chaz 2.0 Apr 13 '12 at 14:48