This tag is for set theory topics typically studied at the advanced undergraduate or graduate level. These include cofinality, axioms of ZFC, axiom of choice, forcing, set-theoretic independence, large cardinals, models of set theory, ultrafilters, ultrapowers, constructible universe, inner model ...

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77
votes
10answers
3k views

Different kinds of infinities?

Can someone explain to me how there can be different kinds of infinities? I was reading "the man who loved only numbers" and came across the concept of countable and uncountable infinities, but ...
61
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 ...
15
votes
1answer
1k views

About a paper of Zermelo

This about the famous article Zermelo, E., Beweis, daß jede Menge wohlgeordnet werden kann, Math. Ann. 59 (4), 514–516 (1904), available here. Edit: Springer link to the ...
20
votes
1answer
3k views

Cardinality of Borel sigma algebra

It seems it's well known that if a sigma algebra is generated by countably many sets, then the cardinality of it is either finite or $c$ (the cardinality of continuum). But it seems hard to prove it, ...
17
votes
3answers
935 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 ...
34
votes
7answers
2k 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 ...
18
votes
2answers
1k 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
1answer
1k 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 ...
18
votes
7answers
4k views

Difference between a class and a set

I know what a set is. I have no idea what a class is. As best as I can make out, every set is also a class, but a class can be "larger" than any set. (A so-called "proper class".) This obviously ...
19
votes
2answers
1k views

Continuity and the Axiom of Choice

In my introductory Analysis course, we learned two definitions of continuity. $(1)$ A function $f:E \to \mathbb{C}$ is continuous at $a$ if every sequence $(z_n) \in E$ such that $z_n \to a$ ...
14
votes
2answers
658 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}< ...
29
votes
3answers
1k views

Set Theoretic Definition of Numbers

I am reading the book by Goldrei on Classic Set Theory. My question is more of a clarification. It is on if we are overloading symbols in some cases. For instance, when we define $2$ as a natural ...
25
votes
2answers
3k views

Is there a known well ordering of the reals?

So, from what I understand, the axiom of choice is equivalent to the claim that every set can be well ordered. A set is well ordered by a relation, $R$ , if every subset has a least element. My ...
10
votes
1answer
869 views

Nonnegative linear functionals over $l^\infty$

My purpose is a clarification of the role of the axiom of choice in constructing limits for bounded sequences. Namely, we want a linear functional of norm 1 defined on the space of all bounded complex ...
8
votes
1answer
839 views

Polish Spaces and the Hilbert Cube

I've been trying to prove that every Polish Space is homeomorphic to a $G_\delta$ subspace of the Hilbert Cube. There is a hint saying that given a countable dense subset of the Polish space $\{x_n : ...
11
votes
2answers
978 views

Uncountable subset with uncountable complement, without the Axiom of Choice

Let $X$ be a set and consider the collection $\mathcal{A}(X)$ of countable or cocountable subsets of $X$, that is, $E \in \mathcal{A}(X)$ if $E$ is countable or $X-E$ is countable. If $X$ is ...
5
votes
2answers
2k views

Cantor-Bernstein-like theorem: If $f\colon A\to B$ is injection and $g\colon A\to B$ is surjective, can we prove there is a bijection as well?

I've been trying to find this proof: If there exists $f \colon A\to B$ injective and $g \colon A \to B$ surjective, prove there exists $h \colon A \to B$ bijective. I thought of using ...
27
votes
3answers
7k views

First-Order Logic vs. Second-Order Logic

Wikipedia describes the first-order vs. second-order logic as follows: First-order logic uses only variables that range over individuals (elements of the domain of discourse); second-order logic ...
59
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. ...
17
votes
2answers
896 views

The set of ultrafilters on an infinite set

After recently learning about filters and ultrafilters, we looked into further problems and properties. I am having trouble with this one: If $X$ is an infinite set, then the set of all ultrafilters ...
2
votes
2answers
712 views

Do $\omega^\omega=2^{\aleph_0}=\aleph_1$?

As we know, $2^{\aleph_0}$ is a cardinal number, so it is a limit ordinal number. However, it must not be $2^\omega$, since ...
12
votes
1answer
432 views

Do sets, whose power sets have the same cardinality, have the same cardinality?

Is it generally true that if $|P(A)|=|P(B)|$ then $|A|=|B|$? Why? Thanks.
5
votes
3answers
455 views

Finite choice without AC

Can anyone explain how we choose one sock from each of finitely many pairs without the axiom of choice? I mean the following quote: To choose one sock from each of infinitely many pairs of socks ...
23
votes
9answers
2k views

Where to begin with foundations of mathematics

I would like to know more about the foundations of mathematics, but I can't really figure out where it all starts. If I look in a book on axiomatic set theory, then it seems to be assumed that one ...
14
votes
7answers
977 views

Applications of ultrafilters

I'm looking for some interesting applications of ultrafilters and also everything of interest involving ultrafilters. Do you know some applications or interesting things involving ultrafilters? I'm ...
19
votes
6answers
2k views

Foundation for analysis without axiom of choice?

Let's say I consider the Banach–Tarski paradox unacceptable, meaning that I would rather do all my mathematics without using the axiom of choice. As my foundation, I would presumably have to use ZF, ...
7
votes
2answers
421 views

How does (ZFC-Infinity+“There is no infinite set”) compare with PA?

How does (ZFC-Infinity+"There is no infinite set") compare with (first order) PA? Intuitively, neither theory should be more powerful than the other.
4
votes
3answers
364 views

Proving that for infinite $\kappa$, $|[\kappa]^\lambda|=\kappa^\lambda$

Initially assume ZFC. Let $\binom{\kappa}{\lambda}$ denotes $\left|[\kappa]^{\lambda}\right|$ where $[\kappa]^{\lambda}$ is the collection of all subsets of $\kappa$ with cardinality $\lambda$. That ...
16
votes
3answers
628 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
1k views

Infinite Set is Disjoint Union of Two Infinite Sets

A finite set is a set such that there exists a bijection from it to some finite ordinal. An infinite set is a set that is not finite. In ZF, can you prove that every infinite set is the union of two ...
11
votes
1answer
494 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 ...
6
votes
1answer
326 views

Is there a way to define the “size” of an infinite set that takes into account “intuitive” differences between sets?

The usual way to define the "size" of an infinite set is through cardinality, so that e.g. the sets $\{1, 2, 3, 4, \ldots\}$ and $\{0, 1, 2, 3, 4, \ldots\}$ have the same cardinality. However, is this ...
6
votes
3answers
1k views

Unary intersection of the empty set

In MK (Morse-Kelley) set theory life is easy: $\forall X\forall y\left(y\in\bigcap X\leftrightarrow\forall x\left(x\in X\rightarrow y\in x\right)\right)$. If $X=\left\{\right\}$ then $\bigcap X=U$, ...
3
votes
1answer
147 views

Godel's pairing function and proving c = c*c for aleph cardinals

I have a few questions about Godel's pairing function and proving that c = c * c for aleph cardinals. Mostly, though, I'm concerned that most of the proofs I've seen are erroneous, and this concerns ...
76
votes
4answers
5k views

What are the Axiom of Choice and Axiom of Determinacy?

Would someone please explain: What does the Axiom of Choice mean, intuitively? What does the Axiom of Determinancy mean, intuitively, and how does it contradict the Axiom of Choice? as simple ...
23
votes
4answers
1k views

Is there a simple, constructive, 1-1 mapping between the reals and the irrationals?

Is there a simple, constructive, 1-1 mapping between the reals and the irrationals? I know that the Cantor–Bernstein–Schroeder theorem implies the existence of a 1-1 mapping between the reals and the ...
10
votes
4answers
2k 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?
13
votes
5answers
792 views

Is there a statement whose undecidability is undecidable?

We know there are statements that are undecidable/independent of ZFC. Can there be a statement S, such that (ZFC $\not\vdash$ S and ZFC $\not\vdash$ ~S) is undecidable?
11
votes
2answers
675 views

Where is axiom of regularity actually used?

Where is axiom of regularity actually used? Why is it important? Are there some proofs, which are substantially simpler thanks to this axiom? This question was to some extent provoked by Dan ...
7
votes
1answer
320 views

What is $\aleph_0$ powered to $\aleph_0$?

By definition $\aleph_1 = 2 ^{\aleph_0}$. And since $2 < \aleph_0$, then $2^{\aleph_0} = {\aleph_1} \le \aleph_0 ^ {\aleph_0}$. However, I do not know what exactly $\aleph_0 ^ {\aleph_0}$ is or how ...
8
votes
3answers
857 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 ...
7
votes
4answers
3k views

Example of set which contains itself

I am trying to understand Russells's paradox How can a set contain itself? Can you show example of set which is not a set of all sets and it contains itself.
13
votes
2answers
304 views

Does $2^X \cong 2^Y$ imply $X \cong Y$ without assuming the axiom of choice?

A friend of mine told me that $X \cong Y \Rightarrow 2^X \cong 2^Y$ ($X$ and $Y$ being sets), which is very easy to prove, but he was wondering about the converse in ZF, i.e., can one take logarithms? ...
5
votes
2answers
291 views

How to think about ordinal exponentiation?

I'm just trying to understand better how to see $\alpha^{\beta}$ for an arbitrary ordinal. I've already know that one can think about $\alpha . \beta$ as $\langle \alpha \times \beta, AntiLex\rangle$ ...
36
votes
11answers
10k views

Why is “the set of all sets” a paradox?

I've heard of some other paradoxes involving sets (ie, "the set of all sets that do not contain themselves") and I understand how paradoxes arise from them. But this one I do not understand. Why is ...
23
votes
4answers
1k views

Is Banach-Alaoglu equivalent to AC?

The Banach-Alaoglu theorem is well-known. It states that the closed unit ball in the dual space of a normed space is $\text{wk}^*$-compact. The proof relies heavily on Tychonoff's theorem. As I have ...
9
votes
1answer
1k views

Axiom of choice and calculus

I thought many results in calculus need axiom choice. For example, I thought one needs AC to prove that a bounded sequence in the real line has a convergent subsequence. Recently I was taught that one ...
14
votes
1answer
773 views

Proofs given in undergrad degree that need Continuum hypothesis?

Or alternative you need to assume CH is false. I know several proofs that use axiom of choice. Heine Borel theorem is the best example I can think off. Zorns lemma is heavily used in the non ...
20
votes
4answers
1k views

Algebraic closure for $\mathbb{Q}$ or $\mathbb{F}_p$ without Choice?

I know the usual proof of the existence of an algebraic closure for any field using Zorn's Lemma. The answer to this previous question makes it clear that in general, some nonconstructive axiom (not ...
9
votes
2answers
1k views

Set theoretic construction of the natural numbers

I'm trying to tie some loose ends here. My lecturer didn't bother to go into details, so I have to work it out myself. I usually hate to be pedantic, but these questions have been bugging me for a ...