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

Martin Sleziak
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  • I have a sequence of blog posts on this topic. It starts here: http://caicedoteaching.wordpress.com/2009/02/05/580-ii-cardinal-arithmetic/ – Andrés E. Caicedo Apr 13 '12 at 16:38
  • And perhaps the most efficient way of looking at all the relevant posts is to go to http://caicedoteaching.wordpress.com/ and then select category 580 (and scroll down). – Andrés E. Caicedo Apr 13 '12 at 16:39
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    [Pete Clark](http://math.stackexchange.com/users/299/pete-l-clark) has some stuff written up in his [website](http://math.uga.edu/~pete/expositions2012.html) under "Set Theory". Chapter 2 gives the basics. – Arturo Magidin Apr 13 '12 at 18:15

1 Answers1


$\newcommand{\alnul}{\aleph_0}\newcommand{\mfr}[1]{\mathfrak{#1}}\newcommand{\Ra}{\Rightarrow}\newcommand{\card}[1]{\left|#1\right|}\newcommand{\powerset}[1]{\mathcal P(#1)}\newcommand{\Lra}{\Leftrightarrow}\newcommand{\Zobr}[3]{#1\colon#2\to#3}$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 $|B|\leq|A|$ 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 $A\times(B\cup C)=A\times B\cup A\times C$.

  • $b\le c \Ra ab\le ac$

See e.g. Proof of cardinality inequality: $m_1\le m_2$, $k_1\le k_2$ implies $k_1m_1\le k_2m_2$ or Will $\kappa_1, \kappa_2, m$ cardinals. Given $\kappa_1 \leq \kappa_2$. prove: $\kappa_1 \cdot m \leq \kappa_2 \cdot m$

  • $a^2=a\cdot a$

See e.g. this answer

  • $a\le b \Ra a^c\le b^c$

See e.g. this answer.

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

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.

  • $a^{b+c}=a^b\cdot a^c$

See e.g. Let $A,B,C$ be sets, and $B \cap C=\emptyset$. Show $|A^{B \cup C}|=|A^B \times A^C|$ and 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?

  • $(ab)^c=a^c\cdot b^c$

See e.g. Equinumerousity of operations on cardinal numbers and How to prove $|{^A}{(K \times L)}| =_c |{^A}{K} \times {^A}{L}|$?

  • $a^b\le 2^{ab}$

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. 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

  • $\alnul+\alnul=\alnul$

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

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 $b+c=bc=\max\{b,c\}$.

See e.g. How to prove that from "Every infinite cardinal satisfies $a^2=a$" we can prove that $b+c=bc$ for any two infinite cardinals $b,c$?

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

Martin Sleziak
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    I've posted a question at meta about the format of answer I've chosen here. http://meta.math.stackexchange.com/questions/3967/would-organizer-posts-be-useful-wellcome-here – Martin Sleziak Apr 13 '12 at 07:31
  • 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 08:12
  • @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 08:20
  • 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 08:24
  • I wrote that proof: [$a^2=a$](http://math.stackexchange.com/questions/54892/about-a-paper-of-zermelo/54904#54904). Cardinals without choice: [Defining cardinality](http://math.stackexchange.com/questions/53770/defining-cardinality-in-the-absence-of-choice/53771#53771). – Asaf Karagila Apr 13 '12 at 08:25
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    You have set the bar high for future "Overview" questions! – The Chaz 2.0 Apr 13 '12 at 14:48
  • Comparability of cardinals (which, like $a^2=a$, is equivalent to choice) should be here, assuming there's something on math.se to link to. – Chris Eagle May 04 '12 at 15:05
  • Good catch @Chris; thanks. I've added some more stuff about inequality of cardinal numbers and if I find corresponding questions, I'll add links to them, too. I made the question CW, so feel free to improve it. – Martin Sleziak May 04 '12 at 16:40
  • Martin, there was a question whether the cardinals are totally ordered. It should be enough to answer Chris' request. I recall that we both answered and yours was accepted. – Asaf Karagila May 10 '12 at 07:45
  • @Asaf You mean [this question](http://math.stackexchange.com/questions/69774/is-the-class-of-cardinals-totally-ordered/). The question is more about whether AC is needed for this result, so I've added it to the part about AC. – Martin Sleziak May 10 '12 at 07:56
  • Yes. I think that the post can use a reorganization somehow. One day I will sit and do that. – Asaf Karagila May 10 '12 at 08:01
  • @Martin: Do you think it might be time for a good revision and addition to this page? I've seen a lot of questions being re-asked (and in some the answers are superior to previous answers). – Asaf Karagila Jan 05 '13 at 09:20
  • @Asaf Feel free to make any improvements you find useful. – Martin Sleziak Jan 05 '13 at 13:26
  • Hi, I'm just here to say this answer is really good, obvious +1. – YoTengoUnLCD Oct 24 '15 at 05:29