This tag is for questions about cardinals and related topics such as cardinal arithmetics, clubs, stationary sets, cofinality, and principles such as $\lozenge$. Do not confuse with [large-cardinals] which is a technical concept about strong axioms of infinity.

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55
votes
6answers
3k views

Why is $\omega$ the smallest $\infty$?

I am comfortable with the different sizes of infinities and Cantor's "diagonal argument" to prove that the set of all subsets of an infinite set has cardinality strictly greater than the set itself. ...
53
votes
1answer
3k views

How do we know an $ \aleph_1 $ exists at all?

I have two questions, actually. The first is as the title says: how do we know there exists an infinite cardinal such that there exists no other cardinals between it and $ \aleph_0 $? (We would have ...
30
votes
6answers
1k views

Why is the Continuum Hypothesis (not) true?

I'm making my way through Thomas W Hungerfords's seminal text "Abstract Algebra 2nd Edition w/ Sets, Logics and Categories" where he makes the statement that the Continuum Hypothesis (There does not ...
30
votes
7answers
1k views

Refuting the Anti-Cantor Cranks

I occasionally have the opportunity to argue with anti-Cantor cranks, people who for some reason or the other attack the validity of Cantor's diagonalization proof of the uncountability of the real ...
20
votes
3answers
1k views

If the infinite cardinals aleph-null, aleph-two, etc. continue indefinitely, is there any meaning in the idea of aleph-aleph-null?

If the infinite cardinals aleph-null, aleph-two, etc. continue indefinitely, is there any meaning in the idea of aleph-aleph-null? Apologies if this isn't a sensible question, I really don't know too ...
19
votes
4answers
665 views

Does $k+\aleph_0=\mathfrak{c}$ imply $k=\mathfrak{c}$ without the Axiom of Choice?

I'm currently reading a little deeper into the Axiom of Choice, and I'm pleasantly surprised to find it makes the arithmetic of infinite cardinals seem easy. With AC follows the Absorption Law of ...
18
votes
2answers
3k views

Cardinality of set of real continuous functions

The set of all ℝ → ℝ continuous functions is c. How to show that? Is there are bijection between ℝn and the set of continuous functions?
18
votes
2answers
268 views

Does $\mathfrak{(p<q)\land(r<s)}$ imply $\mathfrak{p^r<q^s}$?

Does $\mathfrak{(p<q)\land(r<s)}$ imply $\mathfrak{p^r<q^s}$, where $\mathfrak{ p,q,r,s}$ are cardinal numbers? Is it possible to prove in $\mathsf{ZFC}$ that there is a counterexample?
18
votes
2answers
670 views

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

One of the basic (and frequently used) properties of cardinal exponentiation is that $(a^b)^c=a^{bc}$. What is the proof of this fact? As Arturo pointed out in his comment, in computer science this ...
17
votes
2answers
959 views

For every infinite $S$, $|S|=|S\times S|$ implies the Axiom of choice

how to prove the following conclusion: [for any infinite set $S$,there exists a bijection $f:S\to S \times S$] implies the Axiom of choice. Can you give a proof without the theory of ordinal ...
17
votes
5answers
559 views

Why hasn't GCH become a standard axiom of ZFC?

I've never seen a text that includes GCH in the ZFC axioms. I presume this means that GCH has not achieved widespread acceptance. This seems surprising to me, given that: The cardinal numbers ...
17
votes
4answers
417 views

What do the cosets of $\mathbb{R} / \mathbb{Q}$ look like?

$\newcommand{\R}{\Bbb R}\newcommand{\Q}{\Bbb Q}$ Looking at the group of real numbers under addition $(\R, +)$ it contains the (normal) subgroup of rational numbers $(\Q, +)$. I am wondering how to ...
15
votes
3answers
561 views

For any two sets $A,B$ , $|A|\leq|B|$ or $|B|\leq|A|$

Let $A,B$ be any two sets. I really think that the statement $|A|\leq|B|$ or $|B|\leq|A|$ is true. Formally: $$\forall A\forall B[\,|A|\leq|B| \lor\ |B|\leq|A|\,]$$ If this statement is true, ...
15
votes
1answer
652 views

Medial Limit of Mokobodzki (case of Banach Limit)

A classical Banach limit is very usefull concept for me, but there is a problem with the integration and even with the measurability, this means for a sequence $(f_n)_{n\in \mathbb{N}}$ of measurable ...
15
votes
1answer
281 views

How to formulate continuum hypothesis without the axiom of choice?

Please correct me if I'm wrong but here is what I understand from the theory of cardinal numbers : 1) The definition of $\aleph_1$ makes sense even without choice as $\aleph_1$ is an ordinal number ...
15
votes
1answer
739 views

Cardinality of a Hamel basis

What is the cardinality of a Hamel basis of $\ell_1(\mathbb{R})$? Is it deducible in ZFC that it is seemingly continuum? Does it follow from this that each Banach space of density $\leqslant ...
15
votes
2answers
756 views

Defining cardinality in the absence of choice

Under ZFC we can define cardinality $|A|$ for any set $A$ as $$ |A|=\min\{\alpha\in \operatorname{Ord}: \exists\text{ bijection } A \to \alpha\}. $$ This is because the axiom of choice allows any ...
14
votes
4answers
3k views

Cardinality of the set of all real functions of real variable

How does one compute the cardinality of the set of functions $f:\mathbb{R} \to \mathbb{R}$ (not necessarily continuous)?
14
votes
1answer
1k views

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)?
14
votes
3answers
1k views

The Aleph numbers and infinity in calculus.

I have a fairly fundamental question. What is the difference between infinity as shown by the aleph numbers and the infinity we see in algebra and calculus? Are they interchangeable/transposable in ...
14
votes
4answers
849 views

How many non-isomorphic abelian groups of order $\kappa$ are there for $\kappa$ infinite?

Let $\kappa$ be an infinite cardinal. How many non-isomorphic abelian groups of order (cardinality) $\kappa$ are there? For finite $\kappa,$ we can use the classification theorem and obtain the ...
14
votes
1answer
293 views

Bijection between power sets of sets implies bijection between sets?

Is it true that if $X$ and $Y$ are sets and there is a bijection between $\mathcal{P}(X)$ and $\mathcal{P}(Y)$ then there is a bijection from $X$ to $Y$ ?. I believe this should be obvious, but I ...
14
votes
1answer
284 views

The group of permutations with almost all points fixed is a maximal normal subgroup of the symmetric group.

Let $X$ be an infinite set and let $\operatorname{Sym}(X)$ be the symmetric group of $X.$ Let $N$ denote the set of all permutations $\pi\in\operatorname{Sym}(X),$ such that the complement of the set ...
14
votes
1answer
299 views

Cardinality of a locally compact Hausdorff space without isolated points

I am interested in the following result: Theorem. A locally compact Hausdorff topological space $X$ without isolated points has at least cardinality $\mathfrak{c}$. To prove it, one can find two ...
13
votes
2answers
609 views

Relationship between Continuum Hypothesis and Special Aleph Hypothesis under ZF

Special Aleph Hypothesis AH(0) is the claim $2^{\aleph_0}=\aleph_1$, i.e. there is a bijection from $2^{\aleph_0}$ to $\aleph_1$. Continuum Hypothesis CH is the claim $\aleph_0 \leq \mathfrak{a}< ...
13
votes
1answer
126 views

The preorder of countable order types

Consider the set $\mathcal{O}$ of order types corresponding to all posets of cardinality at most $\aleph_0$. The set $\mathcal{O}$ is a preorder under embeddability of its elements (note that some ...
12
votes
10answers
1k views

Intuitive explanation for how could there be “more” irrational numbers than rational?

I've been told that the rational numbers from zero to one for a countable infinity, while the irrational ones form an uncountable infinity, which is in some sense "larger". But how could that be? ...
12
votes
3answers
556 views

Can the cardinality of continuum exceed all aleph numbers in ZF?

More precisely, is either of the following two statements consistent with ZF: $2^{\aleph_0}\geq\aleph_{\alpha}$ for every ordinal number $\alpha$, $2^{\aleph_0}\leq\aleph_{\alpha}\implies ...
12
votes
1answer
2k views

The cardinality of a countable union of countable sets, without the axiom of choice

One of my homework questions was to prove, from the axioms of ZF only, that a countable union of countable sets does not have cardinality $\aleph_2$. My solution shows that it does not have ...
12
votes
2answers
323 views

What's the difference between saying that there is no cardinal between $\aleph_0$ and $\aleph_1$ as opposed to saying that…

... $\aleph_1$ is the immediate successor of $\aleph_0$? I was reading the wiki article on $\aleph_1$ where a distinction is made between the two. If there's isn't a cardinal between $\aleph_1$ and ...
12
votes
3answers
856 views

The cartesian product $\mathbb{N} \times \mathbb{N}$ is countable

I'm examining a proof I have read that claims to show that the Cartesian product $\mathbb{N} \times \mathbb{N}$ is countable, and as part of this proof, I am looking to show that the given map is ...
11
votes
1answer
614 views

bound on the cardinality of the continuum? I hope not

Suppose we don't believe the continuum hypothesis. Using Von Neumann cardinal assignment (so I guess we believe well-ordering?), is there any "familiar" ordinal number $\alpha$ such that, for ...
11
votes
1answer
439 views

There's non-Aleph transfinite cardinals without the axiom of choice?

I can't find anything on this anywhere. The book I'm largely using at the moment is based around ZFC, so it makes no mention of anything other than the Aleph numbers, but according to Wikipedia on the ...
11
votes
2answers
250 views

Is there an axiom of ZFC expressing that GCH fails as badly as possible?

The GCH axiom basically says that for all infinite cardinal numbers $\kappa$, the number of cardinals lying strictly between $\kappa$ and $2^\kappa$ is as small as possible. Namely, there are none. ...
11
votes
2answers
172 views

The relationship of ${\frak m+m=m}$ to AC

Two simple questions: (Of course ${\frak m}$ denotes a cardinal in the weak sense in the claims below.) Can we prove in ZF that $\aleph_0\le{\frak m\Rightarrow m+m=m}$? If not, what is the ...
11
votes
1answer
176 views

Wimpy powerset function

Define the 'wimpy powerset function' $\mathcal{W} : \mathrm{Set} \rightarrow \mathrm{Set}$ by writing $$\mathcal{W}(B) = \{X \in \mathcal{P}(B) : |X| < |B|\}.$$ A few preliminary observations. ...
11
votes
0answers
361 views

Formulations of Singular Cardinals Hypothesis

I have stumbled on a few different formulations of the Singular Cardinals Hypothesis. The most common are: SCH1: $\quad 2^{cf(\kappa)}<\kappa \ \Longrightarrow \ \kappa^{cf(\kappa)}=\kappa^+$ for ...
10
votes
4answers
306 views

Looking for a problem where one could use a cardinality argument to find a solution.

I would like to find an exercise of the type: Find some $x$ in $A\setminus B$. Solution: since $A$ is uncountable and $B$ is countable such $x$ exists...
10
votes
2answers
2k views

What's “the catch” in this question?

I am solving old exam questions and I came across this question: Let $\langle A_n \mid n < \omega\rangle$ disjoint sets such that $\bigcup_{n < \omega}A_n = \mathbb{R}$. Prove that there ...
10
votes
3answers
476 views

Simple cardinal arithmetic

Prove that if $a^2=a$ for each infinite cardinal $a$ then $b + c = bc$ for any two infinite cardinals $b,c$. I tried $b+c=(b+c)^2=b^2+2bc+c^2=b+2bc+c$, but then I'm stuck there.
10
votes
1answer
462 views

Do you need the Axiom of Choice to accept Cantor's Diagonal Proof?

Math people: It is my understanding that set theorists/logicians compare cardinalities of sets using injections rather than surjections. Wikipedia defines countable sets in terms of injections. ...
10
votes
3answers
293 views

Can sets of cardinality $\aleph_1$ have nonzero measure?

$\aleph_1$ is the cardinality of the countable ordinals. It is the least cardinal number greater than $\aleph_0$, and assuming the continuum hypothesis it's equal to $\mathfrak{c}$, the cardinality of ...
10
votes
2answers
253 views

Uncountable chains

$P(\mathbb N)$ = power set of $\mathbb N$. $A \subset P(\mathbb N)$ is a chain if $a,b \in A \implies$ either $a \subseteq b$ or $ b \subseteq a$ That is we have something like this: $$\ldots a ...
10
votes
2answers
380 views

Cardinality of $H(\kappa)$

Again I have trouble with some exercises in Kunen's set theory. In the following, let $\kappa > \omega$ a cardinal. Then I want to show that 1) $|H(\kappa)| = 2^{<\kappa}$ 2) ...
9
votes
3answers
230 views

Existence of a sequence that has every element of $\mathbb N$ infinite number of times

I was wondering if a sequence that has every element of $\mathbb N$ infinite number of times exists ($\mathbb N$ includes $0$). It feels like it should, but I just have a few doubts. Like, assume ...
9
votes
4answers
259 views

Infinite sets and their Cardinality

(I am a 13 year old so when you answer please don't use things that are TOO hard even though I actually can understand quite complex stuff) I was studying Infinite sets and their cardinality (not in ...
9
votes
1answer
208 views

A “reverse” diagonal argument?

Cantor's diagonal argument can be used to show that a set $S$ is always smaller than its power set $\wp(S)$. The proof works by showing that no function $f : S \rightarrow \wp(S)$ can be surjective ...
8
votes
2answers
1k views

Cardinality of the infinite sets

Consider the following problem: Which of the following sets has the greatest cardinality? A. ${\mathbb R}$ B. The set of all functions from ${\mathbb Z}$ to ${\mathbb Z}$ C. The ...
8
votes
4answers
673 views

What's the cardinality of all sequences with coefficients in an infinite set?

My motivation for asking this question is that a classmate of mine asked me some kind of question that made me think of this one. I can't recall his exact question because he is kind of messy (both ...
8
votes
3answers
699 views

Is the class of cardinals totally ordered?

In a Wikipedia article http://en.wikipedia.org/wiki/Aleph_number#Aleph-one I encountered the following sentence: "If the axiom of choice (AC) is used, it can be proved that the class of cardinal ...