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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7
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2answers
357 views

Relationships between AC, Ultrafilter Lemma/BPIT, Non-measurable sets

How is it possible to reconcile the following... In 1970, Solovay constructed Solovay's model, which shows that it is consistent with standard set theory, excluding uncountable choice, that all ...
6
votes
1answer
432 views

Bourbaki Proof of Zorn's Lemma in Lang's Algebra

Serge Lang in his book Algebra has a nice appendix on set theory at the end of the book. In particular, in paragraph 2, pp. 881-884 he provides a proof of Zorn's Lemma from "other properties of sets ...
6
votes
2answers
217 views

Prove that the statement implies the Axiom of Choice

Prove that the following statement implies the Axiom of Choice: Let $ C $ is a set (of sets) and $ B $ is a set such that for all $ c \in C $, there exists a $ b \in B $ such that $ b \not\in c $. ...
3
votes
3answers
359 views

Defining cardinals without choice

According to Wikipedia if we assume AC we define a cardinals number to be an ordinal that is not in bijection with any smaller ordinal. Without AC, one takes the cardinality of a set $X$ to be the ...
0
votes
1answer
88 views

What is a conservative/intersective function?

I can't find any information on what a conservative or intersective function is. ...
12
votes
1answer
654 views

Strength of the statement “$\mathbb R$ has a Hamel basis over $\mathbb Q$”

I would like to know if there are "interesting" equivalences to the statement "$\mathbb R$ has a Hamel basis over $\mathbb Q$". I am not interested in more general statements, like "every vector space ...
2
votes
1answer
219 views

Bijection Definition Clarification

I was reading A Walk Through Combinatorics: An Introduction to Enumeration And Graph Theory (Miklós Bóna) and came across a definition for bijection: Let $X$ and $Y$ be two finite sets, and let $f ...
3
votes
2answers
177 views

Axiom of Choice and Ascending Chain Condition

Matsumura, in his "Commutative Ring Theory" p. 14 proves that "A partially ordered set $\Gamma$ satisfies the ascending chain condition $\Leftrightarrow$ every nonempty subset of $\Gamma$ has a ...
1
vote
2answers
242 views

Axiom of Choice - Naive Counterexample [duplicate]

Possible Duplicate: Axiom of choice question I know there is a lot of discussion on the axiom of choice and, in fact, I attended once a lecture on it, but I still cannot understand the ...
2
votes
1answer
150 views

Proving properties of enumeration of infinite cardinals

I am doing the following exercise from Just/Weese: where $F$ is defined as follows: (a) I'm not sure whether the following passes as "convince myself": Apparently $F$ is defined for all ...
3
votes
1answer
89 views

If $X=\bigcup_{n\in{\mathbb{N}}}\kappa^n$, is it provable from $ZF$ that $|X|=\kappa$?

My question is the following, if $\kappa$ is an aleph and $F$ is the set of all finite sequences in $\kappa$, then the fact that $|F|=\kappa$ is provable from $ZF$?. This can be proven from $ZF$ for ...
8
votes
1answer
279 views

If $\alpha$ is an indecomposable ordinal, why $\Gamma(\alpha\times{\alpha})=\alpha$?

Greets This is from exercise 3.4 of Thomas Jech's "Set Theory", stated: "Show that $\Gamma(\alpha\times{\alpha})\leq{\omega^{\alpha}}$. Thus $\Gamma(\alpha\times{\alpha})=\alpha$ for all ...
22
votes
1answer
599 views

Model existence for infinitary logics

One of the problems of infinitary logic is that it is possible for compactness to fail in a spectacular way: for example, one can concoct an inconsistent set of axioms whose proper subsets are all ...
2
votes
1answer
249 views

Does every infinite well-ordered set have an initial segment isomorphic to $\mathbb N$?

This is an exercise question from Chapter 2 of A Course in Galois Theory by D.J.H. Garling: Suppose that $(A,\leq)$ is an infinite well-ordered set. Show that there is a unique element $a$ such that ...
5
votes
0answers
105 views

Cardinality of $\mathbb R\setminus\mathbb Q$ without AC [duplicate]

Possible Duplicate: Does $k+\aleph_0=\mathfrak{c}$ imply $k=\mathfrak{c}$ without the Axiom of Choice? Showing that $\mathbb{R}$ and $\mathbb{R}\backslash\mathbb{Q}$ are equinumerous using ...
1
vote
1answer
195 views

Problem 1.30 (The Maximum Principle is equivalent to the axiom of choice) (J.Bell, Boolean-valued Models and Independence Proofs, 3rd edition)

Problem 1.30 (The Maximum Principle is equivalent to the axiom of choice) (i) Let $\{a_i : i ∈ I\} ⊆ B$ satisfy $\bigvee_{i∈I} a_i = 1$. A partition of unity $\{b_i : i ∈ I\}$ in B is called a ...
0
votes
0answers
77 views

Problem 1.45 (Boolean-valued ordinals) (J.Bell, boolean-valued models and indipendence proof, 3rd edition)

Problem 1.45 (Boolean-valued ordinals) (J.Bell, boolean-valued models and indipendence proof, 3rd edition) (i) Show that, for any formula φ(x), $ [[ ∃α .φ(α)]] = \bigvee_α [[φ( \hat{α})]]$ and ...
8
votes
2answers
370 views

Does negation of Axiom of Choice imply symmetry?

It seems that every construction of a model in which the Axiom of Choice fails involves some kind of symmetry. Is there an example of a construction of a model where AC fails but no argument involving ...
4
votes
3answers
209 views

What's the thorny issue on: “If all $S\in \ell $ are nonempty, does it follow that $\prod_{S\in \ell} S$ is nonempty? when $\ell$ is infinite?”

I'm reading Paolo Aluffi's ALGEBRA, Chapter 0. Here he proposes that there's a thorny issue: What is this thorny issue?
4
votes
2answers
262 views

How to construct $\{\{\{…\}\}\}$ in ZF without axiom of foundation

I used to think naively the construction is straightforward, which is, if we add one layer innermost each time, then we could have one that corresponds to $\omega$ in Neumann's representation, which ...
3
votes
1answer
334 views

What does AFA , “Every graph has a unique decoration” mean?

All stuff is from Page 1 - 6, Non-Well-Founded Sets, Peter Aczel, which can be found here. Here a GRAPH will consist of a set of NODES and a set of EDGES, each edge being an ordered pair $(n, n')$ ...
4
votes
2answers
175 views

Modal theorems valid in a set theory model

This is the question i would like to discuss, properly stated. Given a model $M$ for a collection of set theory axioms (ZFC, for example), list all basic modal formulas $\phi$ such that $M\Vdash ...
2
votes
2answers
317 views

What does Russell mean when he defines the “Posterity… with respect to the immediate predecessor”?

The the Introduction to Mathematical Philosophy, Russell defines the "posterity" of a given number with respect to the relation "immediate predecessor" as all those terms that belong to every ...
1
vote
2answers
116 views

Infinite union of internal sets not internal

This is homework problem. I need to give an example of internal sets $A_n \subset \mathbb{R}^*$ for which the union $\bigcup _{n=i}^\infty A_n $ is not internal. Also, this whole internal set ...
2
votes
2answers
158 views

Why $\mathrm{rank}(x^y) < \alpha+\omega$, if $x$, $y$ have rank $\le$ $\alpha$?

This question is from Set Theory, Jech(2006), Page 70, 6.5. Rank function is defined as on Page 64: $V_0=\emptyset$, $V_{\alpha+1}=P(V_{\alpha})$, $V_{\alpha}=\bigcup_{\beta<\alpha}V_\beta$, ...
2
votes
2answers
438 views

Using Zorn's Lemma

Background: I am trying to use Zorn's lemma to show the existence of ultrafilters containing an arbitrary filter on a set $X$. My argument goes as follows: Let $\mathcal{F}_0$ be a filter on $X$. If ...
3
votes
1answer
44 views

Proof of every cofinal subclass of $\mathbf{ON}$ is proper

Can you please tell me if my proof of the following claim is correct? Thank you! Claim: Every cofinal subclass of $\mathbf{ON}$ is proper. Proof: Let $A \subset \mathbf{ON}$ be cofinal. Assume $A$ ...
1
vote
1answer
80 views

Operations and relations

To what extent do operations and relations overlap? Is there some more general structure that encompasses both of these things? Thanks
8
votes
2answers
1k views

Simple and intuitive example for Zorns Lemma

Do you know any example that demonstrated Zorn's Lemma simple and intuitive? I know some applications of it, but in these proof I find the application of Zorn Lemma not very intuitive.
20
votes
6answers
5k views

Textbooks on set theory

I want to do a survey of textbooks in set theory. Amazon returns 3582 books for the keywords "set theory". A small somewhat random selection with number of references in Google scholar is the ...
14
votes
2answers
496 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 ...
2
votes
1answer
57 views

Axiom of choice on function [duplicate]

Possible Duplicate: Using a choice function to find an inverse for $F\colon A\to P(B)$ Let $F:A \rightarrow \mathcal P (B)$ be arbitary functions which covers $B$. Use AC to show there is ...
0
votes
1answer
50 views

How to Understand Collection Principle in the Form of First-order Predicate Calulus

On Page 65, Set Theory, Jech(2006), Collection Principle is formulated as follows: $\forall{X}\exists{Y}(\forall{u}\in{X})[\exists{v} \psi(u, v, p) \to(\exists{v\in{Y}}) \psi(u, v, p)]$ ($p$ is ...
3
votes
1answer
91 views

Ordinal arithmetic washing line

Let $\eta$ be the order type of $\mathbb{Q}$. I'm trying to calculate $(1+ \eta) \cdot (\eta + 1)$ and $(\eta + 1) \cdot (1+ \eta)$. So for the first one I'm thinking that you just do this $(1+\eta) ...
6
votes
2answers
148 views

Question about trees and generalizing the Principle of Dependent Choices.

One form of the Principle of Dependent Choices is that for any tree $T$ of height $\omega$ such that every node of $T$ has a successor, there is a branch of $T$ of length $\omega$. In this post, I ...
3
votes
2answers
219 views

Ordinal arithmetic

So I'm having trouble understanding ordinal arithmetic. So if you have $\omega = \bigcup \{ n | n\in\mathbb{N}\}$ How is this defined $\omega^2$ as in the notes I'm reading it has $\omega^2 = ...
3
votes
2answers
130 views

Questions about generalizations of the Principle of Dependent Choices

I've encountered two different generalizations of the Principle of Dependent Choices (DC) to initial ordinals $\kappa>\omega$. First, some notation. By $A\prec B$, I indicate that $A$ is injectable ...
2
votes
1answer
84 views

Proof that recursive construction is equivalent to dependent choice

I am trying to prove the following: In Zf, dependent choice (DC) and recursive construction (RCW) are equivalent where and (RCW): Can you tell me if my proof in one direction is correct and ...
7
votes
1answer
197 views

Why can countable recursive constructions not be done using countable choice?

Why can countable recursive constructions not be done using countable choice? For example, replace $Z$ by $\omega$ in the following theorem Why can't one prove it using countable choice? (The proof ...
5
votes
2answers
271 views

Trying understand a move in Cohen's proof of the independence of the continuum hypothesis

I've read a few different presentations of Cohen's proof. All of them (that I've seen) eventually make a move where a Cartesian product (call it CP) between the (M-form of) $\aleph_2$ and $\aleph_0$ ...
13
votes
3answers
1k views

Every Number is Describable?

Loosely, a number is describable if it can be unambiguously defined by a finite string over a finite alphabet. Numbers such as $\frac{1}{3}$, $\sum_{n=0}^\infty \frac{1}{n!}$, and "the ratio of ...
7
votes
1answer
256 views

Is Zermelo set theory finitely axiomatizable?

I know that ZF is not finitely axiomatizable, but what about Z (i.e. ZF without Replacement)?
2
votes
1answer
76 views

A formula that is upwards absolute but not downwards absolute

I know that any existential formula is upwards absolute (an any universal formula is downwards absolute) but I was looking for an example of a formula that is upwards absolute but not downwards ...
2
votes
1answer
154 views

question on formulations of Generalized Continuum Hypothesis and Singular Cardinal Hypothesis

I hope this is not a silly question(well, not too silly, I hope). After all, a relevent question at a deeper level is already out there, even though it seems the solution is missing. Why not ...
4
votes
1answer
279 views

Adding Cohen reals one at a time

We know that if we start with a ctm $\mathbb{B}$ and force with the poset of finite functions from $\omega$ to $2$, we add a single Cohen real. We also know that if we force with the poset $\mathbb{P} ...
9
votes
4answers
1k views

What would happen if ZFC were found to be inconsistent?

If, one fine day, someone found a contradiction in ZFC (or even ZF), what implications would such an event have for mathematicians? Is there currently any backup axiomatic system on par with ZFC that ...
7
votes
2answers
324 views

What fragment of ZFC do we need to prove Zorn's lemma?

It is extremely well-known that Zorn's lemma is a theorem of ZFC. My interest is in a certain finitely-axiomatisable fragment of ZFC, sometimes called RZC (restricted Zermelo with choice) or ZBQC. The ...
4
votes
1answer
366 views

An exercise from Levy's Basic Set Theory (Exercise 3.17)

Exercise 3.17 on page 136 asks to prove the Milner-Rado paradox. Levy gives hints for two different proofs. I have no problem with the first one but I do not understand the ordering he mentions in ...
6
votes
1answer
78 views

Does $\beta \mathbb N$ embed into $\beta \mathbb N \setminus \mathbb N$?

Is there a clopen subset of $\beta \mathbb N \setminus \mathbb N$ homeomorphic to $\beta \mathbb N$? If so, is there any plausible description of any such a subspace?
4
votes
1answer
183 views

Always win without a winning strategy

On Page 141, Axiom of Choice, Herrlich(2006) Show that if in a game of the form $G(1, X_1, Y_1, A)$, the first player has no winning strategy, then the second player can always win, even ...