Question

Prove that set $C=\{\alpha<\omega_1:f|\alpha:\alpha\xrightarrow{\rm 1:1,onto}\alpha\times\alpha\}$ is closed and unbounded. Alternative solution.

score 50 · Answer 1 · 2018-01-26 12:07:43

Prove that set $C=\{\alpha<\omega_1:f|\alpha:\alpha\xrightarrow{\rm 1:1,onto}\alpha\times\alpha\}$ is closed and unbounded. Alternative solution.

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123

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score 494 · Answer 7 · 2026-04-15 14:49:48

494

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score 270 · Answer 11 · 2026-04-15 14:40:56

270

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Show that $\aleph_{\alpha}^{\aleph_{1}} = \aleph_{\alpha}^{\aleph_{0}}\cdot 2^{\aleph_1}$

Published on 15 Apr 2026 - 14:40

score 255 · Answer 12 · 2026-04-14 08:26:55

255

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score 105 · Answer 13 · 2026-04-13 18:16:04

105

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score 189 · Answer 14 · 2026-04-11 22:22:24

189

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score 581 · Answer 15 · 2026-04-11 08:27:32

581

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List Question

Prove that set $C=\{\alpha<\omega_1:f|\alpha:\alpha\xrightarrow{\rm 1:1,onto}\alpha\times\alpha\}$ is closed and unbounded. Alternative solution.

Why doesn't the independence of the continuum hypothesis immediately imply that ZFC is unsatisfactory?

How to show that MA implies $\mathfrak{d}=\mathfrak{c}$?

The relationship between large cardinals and program termination

Incompleteness Theorem gives a contradiction?

Does there exist a weakly increasing cofinal function $\kappa \to \kappa$ strictly below the diagonal?

Why such stark contrast between the approach to the continuum hypothesis in set theory and the approach to the parallel postulate in geometry?

Countable Chain Condition

Exercises for continuum hypothesis and forcing

Formally what is an unlabeled graph? I have no problem defining labeled graphs with set theory, but can't do the same here.

Show that $\aleph_{\alpha}^{\aleph_{1}} = \aleph_{\alpha}^{\aleph_{0}}\cdot 2^{\aleph_1}$

Target of a superstrong embedding

Prove that if A is countable, then so is $cl_f(A)$.

Is a random intermediate cardinality set nonmeasurable with probability one?

Defining first order logic quantifiers without sets

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