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Questions tagged [set-theory]

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 theory, definability, infinite combinatorics, transfinite hierarchies; etc. More elementary questions should use the "elementary-set-theory" tag instead.

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References for Axiom of Choice and other Axioms of Set Theory [on hold]

I need references regarding the axiom of choice, variations of the axiom of choice, and other set theory axioms which imply or are consequences of some choice axioms. Some or all of these sources ...
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28 views

Cardinality of an uncountable set after union with another uncountable set, that has smaller cardinality.

When $A$ is uncountable and $B$ is a countable set, $|A\setminus B|=|A|$. How can I prove (or disprove) that $|A\setminus a|=|A|$ or $|A\cup a|=|A|$, where $A, a$ are uncountable sets and $|A|>|a|$ ...
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1answer
61 views

Does the Cantor set have the cardinality of the continuum?

I saw somewhere that there are no sets between $\mathbb Q$ and $\mathbb R$ in the sense that there are no set $S\subset \mathbb R$ s.t. $|\mathbb Q|<|S|$ but $|S|<|\mathbb R|$, i.e. all set $S\...
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0answers
13 views

Prove the comparability theorem for well ordered sets in Naive Set Theory - Halmos

I extracted from here to ask a question~ In Naive Set Theory, Halmos phrases the comparability theorem as follows: The assertion is that if $\langle X, \leqslant_X \rangle$ and $\langle Y, \...
1
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1answer
18 views

Organizing proofs by induction over the ordinals

I'm just learning transfinite induction over the ordinals, and I'm finding it a bit difficult to organize proofs. There seems to be several ways to organize almost any proof, but only one of which ...
1
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1answer
22 views

Question regarding algebra-valued models for set theory

Suppose you have the following algebra $4={1, a, b, 0}$ such that $0≤b≤a≤1$. Now we define the model $V^A$ by transfinite recursion. Conjunction and Disjunction are defined as usually as max and min. ...
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1answer
32 views

is it always possible to make a converging sequence to a limit point?

X is a topology; p $\notin$ A $\subset$ X; if p is a limit point of A; that is, every [open subset of X containing p] has non-empty intersection with A, then does there always exist a sequence of A ...
3
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1answer
50 views

Does a $\kappa$-Kurepa tree imply we have a slim $\kappa$-Kurepa tree?

Take a $\kappa$-Kurepa tree to be a tree with more than $\kappa$ branches, of height $\kappa$, each level having cardinality less than $\kappa$ A slim $\kappa$-Kurepa tree is the same but the ...
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1answer
39 views

Does a von Neumann universe $V$ satisfy the $\omega$-global covering property for $V$ itself?

$V$ is an inner model of itself.(ZF(C): Model or Inner Model) Therefore the following theorem, Bukovský's theorem is satisfied: Let N ⊆ M be an inner model. The following are equivalent: 1. N ...
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0answers
25 views

forcing by ${\omega}$-c.c. $\mathbb{P}$ [duplicate]

For example cohen forcing use partial order $\mathbb{P}$ that has ${\omega_1}$-c.c.(c.c.c.). Question: How does forcing by $\mathbb{P}$ that has ${\omega}$-c.c. turn out? Thanks.
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1answer
47 views

Combinations of Lebesgue measurability, the property of Baire and the perfect set property

Lebesgue measurability (LM), the property of Baire (BP) and the perfect set property (PSP) are probably the most prominent among all the regularity properties of sets of reals. Such a set can either ...
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102 views

Does every effectively axiomatizable first-order theory have a finitely axiomatizable conservative extension?

There's a famous theorem (due to Montague) that states that if $\sf ZFC$ is consistent then it cannot be finitely axiomatized. However $\sf NBG$ set theory is a conservative extension of $\sf ZFC$ ...
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0answers
32 views

Reference: Euclidean Plane is not union of disjoint circles

Recently I came across a post over at MO (https://mathoverflow.net/questions/162324/covering-the-space-by-disjoint-unit-circles) that claimed Sierpinski proved that $\mathbb{R}^2$ cannot be expressed ...
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30 views

Cofinality greater than $\omega$ [on hold]

If the cofinality of k is greater than $\omega$, then is it true that $k^{\omega}=k$?
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28 views

A normal filter including the tail sets is $\kappa$-complete

I need to show that if $\cal F$ is a normal filter on a regular uncountable cardinal $\kappa$, and if $\cal F$ contains all tail sets, i.e. all $$ C_\alpha=\{\beta \ : \ \alpha<\beta<\kappa\...
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1answer
31 views

order type of club - cofinality

With $ot(C)$ being the order-type of the club $C$, and $cof(\alpha)$ the cofinality of the ordinal $\alpha$ Show that for every limit ordinal $\alpha$ there is a club $C\subseteq \alpha$ such that $...
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1answer
33 views

Fodor theorem on ultrafilter

Reading some materials on set theory, I know the generalisation of Fodor's theorem: A filter $\cal F$ on a regular uncountable cardinal $\kappa$ is normal if and only if for every regressive ...
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1answer
44 views

Given $S \hookrightarrow T$ construct $U ≈ T$ disjoint from $S$ in Z set theory?

I was recently thinking about the fact (in ZFC) that, given a (first-order) structure $A$ that embeds into another structure $B$, there is some structure $C$ isomorphic to $B$ such that the domain of $...
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2answers
52 views

How can I prove that if $\alpha$ is an ordinal, then there is an initial ordinal $\kappa$ such that $|\alpha|=|\kappa|$?

I'm having trouble understanding initial ordinals. In particular, I can't prove a seemingly trivial theorem about them. Def: An ordinal $\kappa$ is an initial ordinal iff $\forall \delta < \kappa \...
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1answer
34 views

Does the Axiom schema of Replacement imply the Axiom of Infinity? [duplicate]

The axiom of infinity says that the set of natural numbers exists, while the axiom of replacement says that if an object (a member of a set) exists, then all definable mappings of that object yield ...
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1answer
48 views

Are these two extensionality-axioms equivalent?

Let $\epsilon$ be a binary relation on a set $U$. A subset $A \subseteq U$ is called $\epsilon$-transitive iff $$a \mathrel{\epsilon} b \wedge b \in A \Rightarrow a \in A$$ for all $a,b \in U$. For $...
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1answer
58 views

Problem of cardinal assignment

A weak cardinal assignment is any definite operation on sets $A\mapsto |A|$ which satisfies (C1) and (C3), and it is a strong cardinal assignment if it also satisfies (C2). The cardinal numbers (...
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0answers
47 views

Why use transfinite induction?

Why use transfinite induction to prove? I think the inclusion relation is trivial by transfinite recursion. $\mathbf{11.B}$ The Borel Hierarchy Assume now that $X$ is metrizable, so that every ...
8
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0answers
48 views

Natural examples of “set-theorems” provable in MK but not ZFC and NBG

NBG set theory proves nothing additional about sets relative to ZFC. MK set theory, on the other hand, is stronger, meaning it should prove statements about sets that NBG and ZFC don't. Are there ...
2
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0answers
62 views

Semi-rigid boolean algebras

A forcing construction I'm trying to do seems to require a complete atomless boolean algebra (used as a forcing poset) that is "semi-rigid" in the sense defined below. I'm wondering if anyone has ...
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0answers
33 views

Finite axiomatization of second-order NBG

First-order NBG set theory is finitely axiomatizable. The proof of this basically shows that the axiom schemas, in the presence of the other axioms, can be reduced to a finite set of cases, and are ...
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0answers
62 views

Inconsistent axioms

As far as I know, no inconsistency is known to descend from the Zermelo-Fraenkel (ZFC) axioms. Question: historically, has a surprising inconsistency been found to descend from any comparably simple, ...
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0answers
39 views

Schroder-Bernstein Theorem using AC [duplicate]

I realize that the point of the Schroder-Bernstein theorem is to not use the axiom of choice (AC), but I keep reading that the theorem is trivial if one assumes AC. And since much of set theory is ...
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0answers
20 views

Support of functions on trees.

We define a tree $T$ to be a poset $(T,\leq)$ such that $\forall x\in T$, the set $\{y\in T|y\leq x \}$ is well-ordered. Consider partial functions $f:T\rightarrow T$ such that the domain of $f$ is ...
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0answers
51 views

Why we care only about $\Pi_1$ parameters in axiom schemes?

There is exist tradition to formulate axiom schemes such as induction, comprehension or replacment in a way like "for each formula $\phi(y,\overline{x})$ with free variables $y$ and $\overline{x}=x_1,....
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0answers
38 views

for any infinite set A, |A| >= |N| [duplicate]

I try to prove that the power of any infinite set must be equal/bigger then the power of the natural א0. I tried to show that for any infinite set there exists a subset of it, that its power is the ...
6
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0answers
108 views

Two well orderings of an infinite cardinal agree on a large set

I've seen this question but I'm having trouble following the proof given. This is an exercise from Kunen: If $\kappa$ is an infinite cardinal and $\triangleleft$ is a well ordering on $\kappa$, then ...
3
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1answer
48 views

$\bigcap I_\gamma$ unbounded if $I_0 \supset \cdots \supset I_{\gamma}\supset \cdots$ are unbounded

Let be $\kappa$ a regular cardinal and $I_0 \supset \cdots \supset I_{\gamma} \cdots$ are unbounded subsets of $\kappa$ for $\gamma < \lambda <\kappa$ where $\lambda$ is limit. I want to show $\...
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1answer
55 views

How to define the class of terms of a formal language?

Suppose we have a language $\mathfrak{L}=\{\bf{C},\bf{P},\bf{F},\#\}$ consisting of constant, predicate, function symbols and an arity symbol for functions, $\#$, with alphabet $\Sigma(\mathfrak{L})$. ...
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1answer
46 views

Why can't the tower number be $\aleph_0$?

A set $A$ is said to be almost contained in a set $B$ if $A\setminus B$ is finite. A sequence $(A_\alpha)_{\alpha<\lambda}$ of infinite subsets of $\mathbb N$ will be called a tower if for every $\...
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1answer
57 views

Codification of a formal language in set theory.

Starting with an arbitrary class of sets $\Gamma$, can you generate a free semigroup $\Gamma^*$ over $\Gamma$ with the group operation of concatenation ($\frown$)? The goal here is to codify a formal ...
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1answer
41 views

How to prove that the set $C$ is unbounded

My textbook Introduction to Set Theory 3rd by Hrbacek and Jech defines relevant concepts as follows: A set $C \subseteq \omega_1$ is closed unbounded if $C$ is unbounded in $\omega_1$ , ...
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1answer
29 views

Particular case of Ramsey's theorem $\mu \rightarrow (\mu)_{\kappa}^{2}$

It's well-know that: If $\kappa$ is a cardinal and $\mu$ is a infinite cardinal, and if we partition $\mu$ into $\kappa$ sets ($\kappa < cof(\mu)$), then one set contains $\mu$ members. In ...
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1answer
71 views

Is this a typo or a misunderstanding? (Fine Structure in the Handbook of Set Theory)

I'm reading through Schinder and Zeman's handbook chapter on fine structure, and encountered an issue trying to understand the proof of Lemma 7.4. The statement of the lemma is as follows Let $M$ ...
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1answer
62 views

Why is the Axiom of Well-ordered Choice not strong enough to prove Zorn's Lemma?

This is based on this question: How strong is the axiom of well-ordered choice? The "axiom of well-ordered choice" says that any transfinitely-indexed family of sets has a choice function. The ...
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1answer
171 views

How strong is the axiom of well-ordered choice?

I sometimes see references to the "Axiom of Well-Ordered Choice," but I'm not sure how strong it is. It states that every well-ordered family of sets has a choice function. By "well-ordered family," ...
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0answers
30 views

Help understanding Well-Ordering Theorem [duplicate]

Forgive me for my lack of formal notation, I haven't taken any classes on set theory, or any advanced math topics for that matter. From my understanding based on the wikipedia entries, a well-ordered ...
2
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1answer
60 views

Is it consistent with NBG that there are two different satisfaction classes that satisfy the Tarski conditions?

So, NBG can not prove that first order set theory has a satisfication class. However, it is consistent with NBG that such a class exists. My question is if it is consistent with NBG that two such ...
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1answer
74 views

Models of ZFC & Club Sets.

I need help with the following problem: Let $\kappa$ be a weakly inaccessible cardinal. Show there exists a closed and unbounded set of $\alpha < \kappa$ such that $L_\alpha\vDash ZFC$. I know ...
3
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1answer
76 views

What does it mean for a number to be independent of ZFC?

Since the definition of the Busy Beaver function by Radó in 1962, an interesting open question has been what [is] the smallest value of $n$ for which $BB(n)$ is independent of ZFC set theory. Source: ...
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1answer
24 views

Cellularity of product space

Let $\{X_s\}_{s\in S}$ be a family of topological space, and $d(X_s)\leq m$, then the cellularity, i.e., the supremum of the cardinalities of all families of pairwise disjoint non-empty open subsets ...
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1answer
377 views

Connections between Algebraic Topology and Set Theory

(Co) Homology functors are dependent on the homotopy type of the objects they act on and so a lot of results only care about the "loose" classification of spaces (including the use of co-final spectra ...
0
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1answer
38 views

Formalising the equivalence of a (countably) infinite chain of “or”s and exists

I have a countably infinite set $A$ with elements $\{a_0, a_1, ... \}$. I've also been given $P(n) := (b = a_n) \lor P(n+1)$ and that $P(0)$ is true. I could expand this out to a chain of "ors" for ...
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1answer
62 views

Prove that the set of powers of a set, $\{ A^n : n \in \mathbb{N} \}$ exists by using ZFC axioms (without replacement).

I need to prove that the set $\{A^n : n \in \mathbb{N} \}$ using the ZFC axioms (without Replacement). My (rough) plan would be to construct some set containing "more" sets than necessary, then use ...
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1answer
33 views

Example of infinite partition of a set $A$, with no elements in ultrafilter on $A$

Let $\mathcal{F}$ be an ultrafilter on a set $A$. It is easy to show that for every finite partition $\{X_n : n<m\}$ of $A$, there exists some $n<m$ such that $X_n\in\mathcal{F}$. Similarly if ...