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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2
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1answer
38 views

If $A \subset \mathcal{N}^2$ is a $\mathbf{\Sigma}^0_\alpha$ set, then $\{x : (x,x) \in A\}$ is also $\mathbf{\Sigma}^0_\alpha$.

This is the boldface Borel hierarchy on Baire space. Jech states this with a "clearly". What am I missing that makes the statement completely obvious? I clearly have zero intuition for this material....
4
votes
1answer
141 views

$\vDash \varphi$ iff $\| \varphi \|_A =1$ for every boolean valued structure $A$

In the book Axiomatic Set Theory (Takeuti, G; Zaring, W.M. - 1973) the theorem 6.4 states that if $\varphi$ is a closed formula of a given language then it is satisfied in every boolean valued ...
1
vote
0answers
63 views

Enlarging the continuum with $\sigma$-centered forcing

How large can we force the continuum to be if we force with a $\sigma$-centered forcing notion? References to texts discussing the subject would be much appreciated. [A forcing notion $P$ is called ...
1
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3answers
53 views

Ordinals and Countability

I've been watching Prof Su's (Harvey Mudd College) incredible lecture on Ordinals and Transfinite Induction, and I have a few related questions. He starts by drawing and examining three sets of dots, ...
5
votes
0answers
87 views

Measure on $\omega$ defined in the generic extension by an atomless measure algebra is atomless

Work in Cantor space with standard probability measure $m$. Suppose we are given a sequence of measurable sets $\bar{A}=\langle A_n : n\in \omega\rangle$ and a non-principal ultrafilter $U$ and the ...
1
vote
1answer
51 views

Infinite Spectrum Problem

Let us work in a class theory like NBG. For a given first order sentence $\phi$ define $\infty\text{-spectrum}(\phi)$ to be the class of all cardinal numbers $\kappa$ for which there is a model $M$ ...
0
votes
2answers
61 views

I don't understand the axiom schema of separation.

I understand that the axiom schema of separation should assert the existence of a set $y$, subset of a set $z$, where $y=\{x\in z:\varphi \,x\}$ (with $\varphi$ some formula). In the book I'm ...
6
votes
1answer
123 views

How can mathematics work in wildly different set theories?

There are several set theories, e.g. ZFC and NF, which often have different axioms or are even outright contradictory. And yet most of other branches of mathematics, e.g. abstract algebra or topology, ...
7
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2answers
216 views

Kunen exercise III.8.21

Let $f: \omega_1\to \mathbb{R}$ be one-one. Let $g:[\omega_1]^2\to 2$ be such that for any $\alpha<\beta<\omega_1$, $g(\{\alpha, \beta\})$ is $0$ when $f(\alpha)<f(\beta)$, and $1$ otherwise. ...
0
votes
2answers
53 views

Existence of how many sets is asserted by the axiom of choice in this case?

Applying the axiom of choice to $\{\{1,2\}, \{3,4\}, \{5,6\},\ldots\}$, does only one choice set necessarily exist, or all of the $2^{\aleph_0}$ I "could have" chosen? Or something in between? It ...
-6
votes
2answers
84 views

return of Skolem's Paradox? [closed]

The Lowenheim-Skolem theorem means that if ZFC has a model, it has a countable model. However, I think we can show that there aren't any countable models. The overview is that all the elements of ...
2
votes
1answer
42 views

GCH is preserved when forcing with $Fn(\lambda,\kappa)$.

Given a countable transitive model $M$ where $GCH$ holds it is an exercise from Kunen's book to show that GCH also holds in $M[G]$ when $G$ is a $P-$generic filter over $M$, and $P=Fn(\lambda,\kappa)$ ...
0
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0answers
34 views

Cantor normal form to compute sums and products of ordinals: $\omega^{\beta} c+\omega^{\beta'} c' = \omega^{\beta'}c'$ if $\beta'>\beta$

From Wikipedia on Ordinal arithmetic: The Cantor normal form also allows us to compute sums and products of ordinals: to compute the sum, for example, one need merely know that $$\omega^{\...
3
votes
1answer
37 views

Is BPIT equivalent to some ordering principle?

Working in $\mathsf{ZF}$, is $\mathsf{BPIT}$ (Boolean Prime Ideal Theorem) equivalent to some statement of the form "every set can be ***ly ordered"? I know that $\mathsf{BPIT}$ implies that every set ...
4
votes
2answers
72 views

Is it provable in $ZFC$ that if $V_\kappa\vDash ZFC$, then $\kappa$ is strongly inacessible?

The other direction of this implication is pretty obvious, but I'm having a hard time seeing why this direction might be true. I suspect that it isn't, but part of my suspicion comes from my inability ...
2
votes
1answer
36 views

Is there an agreed upon convention for naming ZFC+Large Cardinal Axioms?

Is there an agreed upon convention in general for what to name ZFC+[Large Cardinal Axiom]? Or would one have to state explicitly which axiom was being added? To explain what I mean, note that anyone ...
2
votes
1answer
47 views

$\kappa$-closed forcing preserves stationary sets.

Let's take an uncountable cardinal $\kappa$ which is regular inside the ground model $M$ and $\mathbb{P}\in M$ a forcing notion which is $\kappa-$closed in $M$. I'm trying to prove that every ...
3
votes
1answer
91 views

Ultraproduct with no long descending sequence

I have a countably infinite well-ordered structure $M$ (over a countable language if it helps), and an uncountable regular cardinal $κ$, and I wanted to construct an elementarily equivalent structure ...
0
votes
1answer
99 views

Does the regularity of $\omega_{\alpha+1}$ need Axiom of Choice?

Many books indicate yes to this question. However, I found the only lemma they claim to use AC is the following statement: If $\{A_i\}_{i\in I}$ is a family of sets, then $|\bigcup_{i\in I}A_i|\leq|I|...
1
vote
0answers
44 views

Countable Transitive model where $\exists A\subset \omega_1\;(L[A]\vDash\, \neg CH)$

It is well known that for every subset $A\subset \omega_1$ if $V=L[A]$ then $L[A]\vDash GCH$. In particular $L\vDash \exists A\subset \omega_1\,(L[A]\vDash\, GCH)$. Nonetheless, it is also consistent ...
1
vote
1answer
43 views

Explicit indexing of countable ordinals by sequences of integers?

What I really want is this: A sequence $P_0$, $P_1$,... such that each $P_n$ is a countable partition of $\omega_1$, $P_{n+1}$ is a refinement of $P_n$, and such that if $A_n\in P_n$ for all $n$ and $...
1
vote
2answers
42 views

Question regarding countable ordinals

Feel free to suggest a better title. We're going to regard $0$ as a limit ordinal, as people sometimes do. Let $L$ be the set of countable limit ordinals. If $\alpha\in\omega_1$ there exist a unique ...
4
votes
1answer
82 views

Well-ordering of sets of cardinal numbers

Proposition For every cardinal number $m$ there is a definite next larger cardinal number. This proposition is proved on page 136 of "Proofs from the Book" using the fact that any set of ordinal ...
3
votes
4answers
167 views

Is $\omega-1$ finite?

I saw some videos and read some stuff about ordinals, and it came to me that $\omega-1$ should be finite. My logic is that $\omega$ is the smallest transfinite number, so $\omega-1$ should be finite.....
-2
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1answer
29 views

Question about Hartogs' theorem proof [closed]

Is it possible to prove this theorem without the use the replacement axiom??
3
votes
2answers
38 views

“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
91 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
vote
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
vote
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
70 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
votes
1answer
58 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
72 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
votes
1answer
101 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
95 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
vote
1answer
34 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 ...
7
votes
0answers
71 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
94 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
52 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
63 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
votes
1answer
57 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
69 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
votes
0answers
13 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
108 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 ...