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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71
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 ...
55
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
898 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 ...
16
votes
2answers
822 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 ...
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 ...
32
votes
6answers
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 ...
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
6answers
3k 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 ...
14
votes
2answers
630 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}< ...
26
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 ...
23
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 ...
18
votes
2answers
972 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 $z_n \to a$ ...
15
votes
1answer
858 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 ...
9
votes
1answer
828 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
768 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 : ...
5
votes
2answers
1k 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
6k 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 ...
16
votes
2answers
837 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
611 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 ...
11
votes
2answers
867 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 ...
12
votes
1answer
409 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.
4
votes
3answers
431 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 ...
21
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 ...
13
votes
7answers
916 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 ...
15
votes
1answer
945 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
469 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
2answers
379 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.
6
votes
1answer
300 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 ...
69
votes
4answers
4k 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 ...
57
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. ...
13
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, ...
9
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?
4
votes
3answers
339 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 ...
15
votes
3answers
594 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, ...
8
votes
3answers
768 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
1answer
279 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 ...
13
votes
2answers
285 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? ...
-3
votes
1answer
303 views

Question about Cantors Diagonal Argument [closed]

Lets be honnest I don't understand cantors diagonal argument. http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument While I can understand that this proof proves that you cannot make the list of ...
23
votes
4answers
952 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 ...
14
votes
1answer
722 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 ...
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 ...
18
votes
4answers
941 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 ...
12
votes
4answers
736 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?
12
votes
2answers
669 views

Is there an algebraic homomorphism between two Banach algebras which is not continuous?

According to wikipedia, you need the Axiom of Choice to find a discontinuous map between two Banach spaces. Does this procedure also apply for Banach algebras yielding a discontinuous multiplicative ...
11
votes
2answers
395 views

Any good decomposition theorems for total orders?

I'm learning about set theory and partial orders, well orders, total orders, etc. I'm curious to know of any good decomposition theorems that apply to all (or very general types of) total orders. ...
7
votes
1answer
615 views

Uncountable ordinals without power set axiom

Assume $M$ is a set, in which all axioms of $ZF - P + (V=L)$ hold. Does then $M$ believe that there exists an uncountable ordinal? I mean, why should the class of all countable ordinal numbers be a ...
7
votes
3answers
232 views

What can I do with proper classes?

There are standard tricks, constructions and techniques in ZFC when working with proper classes; for instance one can form the cartesian product of a pair of classes without difficulty, or more ...
3
votes
2answers
514 views

How do I choose an element from a non-empty set?

Suppose I have a non-empty set $A$. How do I choose an element $x\in A$? More precisely, I believe I would like to find a formula $P(x,y)$ of ZF such that for every non-empty set $y$ there is ...
7
votes
6answers
1k views

Axiom of Choice and finite sets

So I am relatively familiar with the Axiom of Choice and a few of its equivalent forms (Zorn's Lemma, Surjective implies right invertible, etc.) but I have never actually taken a set theory course. I ...
6
votes
4answers
2k 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.