Questions tagged [cardinals]

This tag is for questions about cardinals and related topics such as cardinal arithmetics, regular cardinals and cofinality. Do not confuse with [large-cardinals] which is a technical concept about strong axioms of infinity.

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Value of the cardinal product $\aleph_1 \cdot \mathfrak{c}$

Suppose we want to know how many well-orderings of the naturals there are. That is, not up to isomorphism, but how many individual ways to well order the naturals there are. It's easy to see that for ...
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Set of various order types of a set

Starting from the cardinal $|\Bbb N| = \aleph_0 = \beth_0$, we can generate a larger cardinal in two ways: Take the set of all subsets, generating the cardinal $\beth_1$ Take the set of all well-...
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A question regarding Brian M. Scott's proof that $\text{cf}(\aleph_{\omega_1})=\omega_1$

$\text{cf}(\aleph_{\omega_1})=\omega_1$ From here, I quote Brian M. Scott's proof: Suppose that $\langle\alpha_n:n\in\omega\rangle$ is an increasing sequence cofinal in $\omega_{\omega_1}$. For ...
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Let $\alpha$ be a limit ordinal which is not a cardinal, and $\kappa=|\alpha|$. Then there exists a bijection from $\kappa$ to $\alpha$, or equivalently, a one-to-one sequence $\langle \alpha_\xi \... 1answer 81 views How can we determine$2^\kappa$for singular$\kappa$assuming that$2^\lambda=\lambda^+$whenever$2^{\operatorname{cof}(\lambda)}<\lambda$? I got an exercise in set theory and can't seem to solve it: If we assume that$2^\lambda = \lambda^+ $holds for every singular cardinal with$2^{\operatorname{cof}(\lambda)}<\lambda$, then how can ... 1answer 80 views If$\alpha$is a limit ordinal, then$\operatorname{cf}(\alpha)\$ is a limit ordinal [duplicate]

In the textbook Introduction to Set Theory by Hrbacek and Jech, Section 9.2, the authors first introduce the definition of increasing sequence of ordinals: Then they introduce cofinality: My ...