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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3
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2answers
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“Proof” that $\text{cof}(\omega_\lambda)<\omega_\lambda$ if $\lambda$ is a nonzero limit ordinal

The following lemma is from this introduction to cardinals. Lemma 2.7. Let $\omega_\alpha$ be a limit cardinal. Then $\alpha$ is a limit ordinal and $\text{cof}(\omega_\alpha)=\text{cof}(\alpha)$. ...
2
votes
3answers
92 views

Proving the class of countable ordinals is closed under ordinal exponentiation in ZF

I managed to prove that given the axiom of choice, the class (or is it a set?) of countable ordinals is closed under exponentiation, since the axiom of choice implies that the countable union of ...
1
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1answer
46 views

For every cardinal $\kappa$, $\kappa^+$ is regular

Again I'm struggling with a proof from this introduction to cardinals. Lemma 2.6. For every cardinal $\kappa$, $\kappa^+$ is regular. Proof. If not, then there would be a cofinal map $f:\...
1
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1answer
37 views

Using the axiom of choice to choose bijections

I couldn't think of a better question title. I am trying to understand the proof of theorem 1.8 in this introduction to cardinals. Theorem 1.8. Let $\kappa\in CARD$. Let $X=\bigcup_{\alpha<\...
3
votes
1answer
72 views

Every $\alpha < \kappa^+$ can be embedded in any interval of $\kappa^{<\omega}$.

Let $\kappa$ be a cardinal. I want to show that every ordinal $\alpha < \kappa^+$ can be embedded (as an order) in any interval of $\kappa^{<\omega}$, ordered lexicographically. This is exactly ...
0
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1answer
60 views

Axiom of Foundation (regularity) implies epsilon induction

I'm trying to understand why epsilon induction is equivalent to foundation, given the other axioms of ZF. In another post, it is shown that epsilon induction implies foundation, and I understand that ...
-2
votes
1answer
73 views

Is ZFC the Minimal Structural Theory that Models R [closed]

Is ZFC minimal or "simplest" in the sense that is it is the structural theory with the minimal number of sub-structures that can model R and its "classic" continuity and completeness properties (maybe ...
1
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1answer
41 views

Regarding Choice in fields outside set theory.

When authors say stuff like The equivalence of continuity and sequential continuity in metric spaces uses(/requires) some version of the axiom of choice. Are they assuming that we are working ...
0
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1answer
103 views

Measure and set theory.

I have read that if we assume the continuum hypothesis then it can be proved or concluded tha there exist a set function μ that has the three following properties: μ(A) is defined for each set A of ...
0
votes
1answer
99 views

Is there a set without a predicate? [closed]

Is there a set that has no predicate that defines it? I limit this question to the pure set theory. It seems there are sets whose members have no common exclusive property and so the only way to ...
0
votes
3answers
46 views

Can an ordinal be limit of a smaller set of ordinals?

The Wikpedia article Regular cardinal contains the following weird sentence: An infinite ordinal $\alpha$ is regular if and only if it is a limit ordinal which is not the limit of a set of smaller ...
1
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1answer
35 views

No uncountable subset can be well-ordered $\iff$ Hartogs ordinal is $\omega_1$

I'm looking at an exercise which starts as follows: Suppose that no uncountable subset of $\mathcal P\omega$ can be well-ordered (equivalently, the Hartogs ordinal $\gamma(\mathcal P\omega)$ is $\...
7
votes
1answer
186 views

Showing a cardinal is regular

I've been thinking about this question, but to no avail and I've got to ask. How to show that for $\kappa\geq\aleph_0,$ $\mu=\min\{\lambda: \kappa^{\lambda} > \kappa\}$ is regular? If I wanted a ...
3
votes
2answers
36 views

Another characterization of the cofinality?

Is it true that $cf(\kappa)=\min \{\lambda:\ \kappa^{\lambda}>\kappa\}$? $cf(\kappa)$ is certainly $\geq$ than that minimum since $\kappa^{cf(\kappa)}>\kappa$, but I don't know how to tackle ...
8
votes
0answers
73 views

Question about a linear order

Let $L$ be the set of countable limit ordinals. For each $\alpha \in L$, let $\langle \alpha_n : n < \omega \rangle$ be a strictly increasing cofinal sequence in $\alpha$. Define a linear order on $...
2
votes
2answers
100 views

What are some applications of large cardinals?

Most mathematicians don't often encounter cardinalities larger than that of the continuum, it seems? What are some results outside of pure set theory/logic that rely on the properties of larger ...
2
votes
1answer
56 views

Is there any well-ordered uncountable set of real numbers under the original ordering?

I know that the usual ordering of $\mathbb R$ is not a well-ordering but is there an uncountable $S\subset \mathbb R$ such that S is well-ordered by $<_\mathbb R$? Intuitively I'd say there is no ...
0
votes
1answer
36 views

Get well-order from total order by choosing finite subsets

Exercise 7.9 in this book: Given that any set has a multiple-choice function, ie a function picking out a nonempty finite subset of each nonempty subset, show that any totally orderable set can be ...
4
votes
1answer
65 views

Does ACH+NJA$(\mathbb{R})$ decide the Suslin problem?

Lets start with a couple of axioms in the language of ZFC. ACH. Anticontinuum Hypothesis. The cardinal $|\mathbb{R}|$ is an aleph fixed points: that is, $$|\mathbb{R}| = \aleph_{|\mathbb{...
0
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1answer
58 views

consistency of ZF\{Power Set axiom}

Assuming consistency of ZF, ZF$\setminus ${Power Set axiom} is consistent. But can we prove consistency of ZF$\setminus ${Power Set axiom} without assuming consistency of ZF.
2
votes
1answer
71 views

Without the Axiom of Choice, $\aleph_0<2^{\aleph_0}$ implies $\aleph_1\le 2^{\aleph_0}$?

Question: In ZF (so AC does not necessarily hold) does the following claim hold? $\aleph_0<2^{\aleph_0}$ implies $\aleph_1\le 2^{\aleph_0}$ This question arose to me when reading the top ...
4
votes
1answer
52 views

An axiom combining Choice and replacement

I am just finishing teaching a course in Set Theory, and was thinking about the ZFC axioms, when the following axiom occurred to me: Axiom: Given a sentence $S$ with free variables among $x,y,A_1,\...
0
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0answers
14 views

Decomposition of set with continuum [duplicate]

Let $E=\bigcup\limits_{n=1}^{\infty}A_n$ and $card(E)=card(\mathbb R)$. Show that there exists $n_0$ such that $card(A_{n_0})=card(\mathbb R)$. Any method, without admitting continuum hypothesis, can ...
3
votes
1answer
77 views

Proof of Kondô-Addison theorem

The proof of the (lightface) Kondô-Addison theorem (aka $\Pi^1_1$ uniformization) that I know goes like this: for a $\Pi^1_1$ set $R \subseteq 2^\omega \times 2^\omega$, define the uniformization of $...
8
votes
2answers
110 views

Can some mathematical objects exist without set theory?

I'm a beginner in undergraduate mathematics, and am studying subjects like real analysis and abstract algebra. It seems that most mathematical objects hinge on set theory. A group is a set equipped ...
1
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1answer
47 views

Does König's inequality require axiom of choice?

König's inequality: If $\kappa_\alpha<\lambda_\alpha$ for all $\alpha<\mu$, then $\sum\kappa_\alpha<\prod\lambda_\alpha$. In order to prove this inequality, we need to provide an injection ...
2
votes
1answer
32 views

Does the cofinality always induce a continuous function?

Let $\alpha$ be a limit ordinal and let $\lambda = cf(\alpha)$ be its cofinality. This means that there exists a strictly increasing $\lambda$-sequence $\langle \alpha_\xi \mid \xi < \lambda \...
3
votes
1answer
65 views

Are the following two definitions of infinite sets logically equivalent?

Dedekind-Infinite Set. A set $S$ is called a Dedekind-Infinite Set if there exists a bijection $f:S\to T$ where $\emptyset \subset T\subset S$. I was wondering if the following definition is ...
3
votes
3answers
63 views

Question about $\aleph$-fixed point

I am working through a proof on cardinals I found and can't reason some of the steps. The proposition is that there is an $\aleph$-fixed point, i.e. there is an ordinal $\alpha$ (which is ...
3
votes
3answers
33 views

Is the equivalence of the ascending chain condition and the maximum condition equivalent to the axiom of dependent choice?

Assuming the axiom of dependent choice, for a partially ordered set $(X,\le)$, the following statements are equivalent: $X$ fulfils the ascending chain condition, i. e. every chain $x_1\le x_2\le\...
6
votes
3answers
254 views

Why is establishing absolute consistency of ZFC impossible?

Why is establishing the absolute consistency of ZFC impossible? What are the fundamental limitations that prohibit us with coming up with a proof? EDIT: This post seems to make the most sense. In ...
0
votes
2answers
39 views

What's the correct form of Axiom of Extensionality?

Different sources report two different forms of the axiom (in which $=$ is considered a primitive notion): 1.${\forall}v_0({\forall}v_1({\forall}v_2((v_2{\in}v_0){\iff}(v_2{\in}v_1)){\iff}(v_0=v_1)))$...
3
votes
4answers
375 views

How to prove that a set exists in ZFC?

This is probably something really trivial, but I don't have any helpful set theory books (nor any library where I could borrow them for that matter) and googling such things as "proving a set exists ...
5
votes
1answer
66 views

$\aleph_1$ almost sure events that almost never all hold

This recent question sparked my curiosity. Is there a family of events $(E_k)_{k \in I}$ such that$\def\pp{\mathbb{P}}$ $\pp(E_k) = 1$ for any $k \in I$ but $\pp( \bigcap_{k \in I} E_k ) = 0$? Clearly ...
4
votes
1answer
142 views

What is the *exact* consistency strength of $MK$?

It's well known that the existence of an inaccessible cardinal proves $Con(MK)$. Joel Hamkins has a nice blog post (http://jdh.hamkins.org/km-implies-conzfc/) that explains what you get out of $MK$, ...
1
vote
1answer
30 views

Does the relationship between Jonsson Cardinals and Jonsson Algebras require the axiom of choice?

Using Skolem functions, one can see in ZFC that a cardinal $\kappa$ is Jonsson iff there are no Jonsson algebras on $\kappa$. (I.e. every algebra of size $\kappa$ has a proper subalgebra of size $\...
0
votes
1answer
78 views

Is predicate a set?

A set can be defined by a predicate. Is the predicate itself a set too? A predicate seems to be a sort of relation and since relations are sets (in set theory), it seems that a predicate is a set ...
1
vote
1answer
71 views

Proving ZFC is consistent

I've heard from a friend that we can actually prove the consistency of ZFC if we assume at least one inaccessible cardinal exists. How is this carried out, precisely? Googling doesn't help and my ...
-4
votes
1answer
44 views

show that there's a linear ordering such that $R \subseteq R^*$ [duplicate]

Show that if $R$ is a partial ordering on a set $A$, then there exists a linear ordering $R^*$ on $A$ such that $R \subseteq R^*$
1
vote
1answer
41 views

Hausdorff Maximal Principle

"Hausdorff's Maximal Principle" says that any partial order P has a maximal chain (chain = linear suborder). It is equivalent to the axiom of choice. If we restrict Hausdorff's Maximal Principle to ...
1
vote
1answer
32 views

Preserving weakly inaccesible cardinals in generic extensions

I'm trying to see that every weakly inaccessible cardinal $\kappa$ in a ctm $M$ remains weakly inaccesible if we force with a forcing which preserves cofinalities (thus, preserves cardinalities). It's ...
1
vote
1answer
45 views

Why Can we work with $M$ model countable transitive model of some finite fragment of $\mathrm{ZFC}$ and why is it exist.?

When we say that let $M$ be a countable transitive model of some finite fragment of $\mathrm{ZFC}$. Why Can we work with $M$ model countable transitive model of some finite fragment of $\mathrm{ZFC}$...
2
votes
1answer
104 views

Regarding the axiom $2^\kappa = 2^{\kappa^+}$ for regular cardinals $\kappa$, and its relationship to a couple of other axioms.

(Take ZFC as background.) The following two statements both follow from GCH: ICF. Injective continuum function. The continuum function (i.e. $\kappa \mapsto 2^\kappa)$ is injective. NJA....
2
votes
1answer
41 views

Do the notations for relative constructible universe and for forcing extention coincide?

We have the notations $L[A]$ for the inner-model constructible relative to some $A$, and the notation $M[G]$ for a generic extention of the model $M$. Do they coincide? That is, if we look at the ...
0
votes
1answer
34 views

Decidability of the surjectivity of any given function in $\mathsf{ZFC}$

Let $X$ and $Y$ be two non-empty sets and let $f:X\to Y$ be an arbitrary function. Let $\mathsf P$ be the statement that $f$ is surjective. That is, $$\mathsf P\equiv[\forall y\in Y\,\exists x\in X:y=...
0
votes
1answer
77 views

Forcing exercise from Kunen's book

I'm new in the study of the forcing method and I having some troubbles to solve some of the exercise from Kunen's book (edition 2013): specifically, problem IV 2.46 from page 271. It says the ...
7
votes
1answer
68 views

Does $\operatorname{card}(X) < \operatorname{card}(Y)$ imply $\operatorname{card}(X^2) < \operatorname{card}(Y^2)$ without choice?

I looked to see if this question was already posted, but did not find anything. Please let me know if this is a duplicate. Assume $X, Y$ are infinite sets such that there is an injection $X \to Y$ ...
2
votes
1answer
69 views

Why can't we quantify over propositional functions in the ZFC set theory?

What's the difference between saying if $P(y)$ is some propositional function, then we can create an axiom ${\forall}z{\exists}x:(y{\in}x{\iff}y{\in}z{\land}P(y)$ and saying ${\forall}P(...
1
vote
2answers
68 views

Does Second-Order Comprehension make second-order ZFC inconsistent due to Russell's Paradox?

When we do set theory, we take our first-order variables to range over all sets. But if we take our second-order variables to range over sets of sets in the range of the first-order variables, then ...
0
votes
0answers
42 views

How can one formulation of the axiom of extension define set equality and the second one not?

I've seen the claim that there are two ways of writing the axiom of extension. The first one is ${\forall}A{\forall}B({\forall}x(x{\in}A{\iff}x{\in}B)){\iff}A=B$. This one supposedly admits $=$ as a ...