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.

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71
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1answer
5k 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 ...
67
votes
6answers
4k 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. ...
43
votes
11answers
3k 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 ...
38
votes
7answers
4k 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 ...
32
votes
8answers
2k views

Is symmetric group on natural numbers countable?

I guess it is too difficult a question to ask about the cardinality of $S_{\mathbb{N}}$ so I would like to ask whether it is countable or not. I tried to prove it is uncountable somewhat mimicking ...
30
votes
1answer
4k 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)?
29
votes
6answers
9k views

Cardinality of set of real continuous functions

The set of all $\mathbb{R\to R}$ continuous functions is $\mathfrak c$. How to show that? Is there any bijection between $\mathbb R^n$ and the set of continuous functions?
28
votes
7answers
3k views

Why do we classify infinities in so many symbols and ideas?

I recently watched a video about different infinities. That there is $\aleph_0$, then $\omega, \omega+1, \ldots 2\omega, \ldots, \omega^2, \ldots, \omega^\omega, \varepsilon_0, \aleph_1, \omega_1, \...
27
votes
3answers
950 views

What is the largest set for which its set of self bijections is countable?

I recently came across a problem which required some knowledge about the self bijections of $\mathbb{N}$, and after looking up how to construct some different bijections I came across the result that ...
24
votes
4answers
914 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 ...
22
votes
2answers
2k 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 numbers....
22
votes
3answers
2k 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 ...
21
votes
2answers
2k views

Cardinality of a Hamel basis of $\ell_1(\mathbb{R})$

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 2^{\...
20
votes
3answers
1k 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 ...
20
votes
2answers
335 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?
20
votes
2answers
1k 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 ...
19
votes
2answers
1k views

Are there number systems corresponding to higher cardinalities than the real numbers?

As most of you know, the set $\omega$ with cardinality $\aleph_0$ corresponds to what we normally know as the natural numbers $\mathbb{N}$, and the set $\mathcal{P}(\omega)$ with cardinality $\aleph_1$...
19
votes
5answers
822 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 ...
19
votes
2answers
373 views

Covering $\mathbb R^2$ with function graphs

Suppose we have a countable family of function graphs (each function is $\mathbb R\to\mathbb R$, not necessary continuous). Obviously, they cannot cover the whole plane $\mathbb R^2$, because they ...
18
votes
10answers
8k 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 form a countable infinity, while the irrational ones form an uncountable infinity, which is in some sense "larger". But how could that be? ...
18
votes
3answers
2k views

How many non-differentiable functions exist?

The size of the set of functions that map $\mathbb{R}\to \mathbb{R}$ equals $(\#\mathbb{R})^{\#\mathbb{R}}$. How many non-differentiable functions are there in this set?
18
votes
3answers
810 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, ...
18
votes
4answers
7k 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)?
18
votes
4answers
3k views

Cardinality of all cardinalities

Let $C = \{0, 1, 2, \ldots, \aleph_0, \aleph_1, \aleph_2, \ldots\}$. What is $\left|C\right|$? Or is it even well-defined?
18
votes
1answer
545 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 (...
18
votes
1answer
3k 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 ...
18
votes
4answers
649 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 ...
17
votes
5answers
1k views

Why study cardinals, ordinals and the like?

Why is the study of infinite cardinals, ordinals and the like so prevalent in set theory and logic? What's so interesting about infinite cardinals beyond $\aleph _0 $ and $\mathfrak{c} $? It seems ...
17
votes
3answers
2k 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 ...
16
votes
1answer
914 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 (...
16
votes
3answers
2k 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 ...
16
votes
1answer
647 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 can'...
16
votes
1answer
442 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 ...
16
votes
1answer
203 views

There's no cardinal $\kappa$ such that $2^\kappa = \aleph_0$

I am trying to prove that there is no cardinal $\kappa$ such that $2^\kappa = \aleph_0$ . My attempt: We suppose it exists. Since $\kappa<2^\kappa$, in particular, $\kappa<\aleph_0$. But ...
15
votes
2answers
793 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}< ...
14
votes
3answers
1k views

What is an example of two sets which cannot be compared?

In set theory, if we do not assume the Axiom of Choice, we cannot prove the Trichotomy Law between cardinals. That is, we cannot prove that for any two sets, there exists an injection from one to the ...
14
votes
1answer
909 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. ...
14
votes
4answers
6k views

Do the real numbers and the complex numbers have the same cardinality?

So it's easy to show that the rationals and the integers have the same size, using everyone's favorite spiral-around-the-grid. Can the approach be extended to say that the set of complex numbers has ...
14
votes
4answers
1k 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
2answers
500 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 $\...
14
votes
3answers
639 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 ...
14
votes
1answer
310 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
2answers
294 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 ...
14
votes
0answers
475 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 ...
13
votes
4answers
839 views

Are there more groups than rings?

It seems pretty clear to me that both of these are at least uncountable (which I think I could prove with some work). It also seems that you should be able to make some diagonal argument about the two,...
13
votes
2answers
374 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 ...
12
votes
1answer
896 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 non-...
12
votes
3answers
854 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 2^{\...
12
votes
1answer
608 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 ...
12
votes
2answers
305 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. ...