# Tagged Questions

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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### “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 $\text{cof}(\omega_\alpha)=\text{cof}(\alpha)$. ...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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 ...
### 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 $X \to Y$ ...