Question

Ordinal arithmetic with multiple brackets

score 101 · Answer 1 · 2018-11-24 18:15:49

101

Views

Ordinal arithmetic with multiple brackets

Published on 24 Nov 2018 - 18:15

score 244 · Answer 2 · 2018-11-24 21:41:51

244

Views

Ordinal multiplication

Published on 24 Nov 2018 - 21:41

score 131 · Answer 3 · 2018-11-25 13:48:32

131

Views

Making sense of a bijection between proper classes?

Published on 25 Nov 2018 - 13:48

score 93 · Answer 4 · 2018-11-28 12:58:17

93

Views

The countable union of countable ordinals is countable, using ZF

Published on 28 Nov 2018 - 12:58

score 109 · Answer 5 · 2018-11-29 15:46:12

109

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$\omega_2$ is not the countable union of countable sets

Published on 29 Nov 2018 - 15:46

score 207 · Answer 6 · 2018-12-01 08:42:17

207

Views

Prove that $|\omega+\omega|=|\omega\cdot\omega|=|\omega^\omega|=|\omega|$

Published on 01 Dec 2018 - 8:42

score 33 · Answer 7 · 2018-12-01 12:43:51

33

Views

Is it true that the class of initial ordinals is proper?

Published on 01 Dec 2018 - 12:43

score 37 · Answer 8 · 2018-12-01 14:03:19

37

Views

Let $n,k\in\omega$. Then $\underbrace{(\omega+k)+(\omega+k)+\ldots+(\omega+k)}_{n\text{ times}}=\omega\cdot n+k$

Published on 01 Dec 2018 - 14:03

score 106 · Answer 9 · 2018-12-02 15:14:47

106

Views

Each well-orderable set $X$ is equipotent to a unique initial ordinal

Published on 02 Dec 2018 - 15:14

score 119 · Answer 10 · 2018-12-05 02:57:52

119

Views

The Hartogs number of $A$ exists for any set $A$

Published on 05 Dec 2018 - 2:57

score 51 · Answer 11 · 2018-12-05 03:09:59

51

Views

Is the function sending each set to its corresponding Hartogs number injective?

Published on 05 Dec 2018 - 3:09

score 399 · Answer 12 · 2026-03-25 01:25:59

399

Views

Is there a "solution" to the ordinal game?

Published on 25 Mar 2026 - 1:25

score 1.7k · Answer 13 · 2018-12-07 02:56:39

1.7k

Views

Axiom of Choice implies Well-Ordering Principle

Published on 07 Dec 2018 - 2:56

score 952 · Answer 14 · 2018-12-07 06:01:23

952

Views

Well-Ordering Principle implies Zorn's Lemma

Published on 07 Dec 2018 - 6:01

score 51 · Answer 15 · 2018-12-07 10:08:53

51

Views

If $(A,\preccurlyeq)$ is a linear ordering such that $|\{y\in A\mid y\preccurlyeq x\}| < \aleph_\gamma$ for all $x\in A$, then $|A|\le\aleph_\gamma$

Published on 07 Dec 2018 - 10:08

List Question

Ordinal arithmetic with multiple brackets

Ordinal multiplication

Making sense of a bijection between proper classes?

The countable union of countable ordinals is countable, using ZF

$\omega_2$ is not the countable union of countable sets

Prove that $|\omega+\omega|=|\omega\cdot\omega|=|\omega^\omega|=|\omega|$

Is it true that the class of initial ordinals is proper?

Let $n,k\in\omega$. Then $\underbrace{(\omega+k)+(\omega+k)+\ldots+(\omega+k)}_{n\text{ times}}=\omega\cdot n+k$

Each well-orderable set $X$ is equipotent to a unique initial ordinal

The Hartogs number of $A$ exists for any set $A$

Is the function sending each set to its corresponding Hartogs number injective?

Is there a "solution" to the ordinal game?

Axiom of Choice implies Well-Ordering Principle

Well-Ordering Principle implies Zorn's Lemma

If $(A,\preccurlyeq)$ is a linear ordering such that $|\{y\in A\mid y\preccurlyeq x\}| < \aleph_\gamma$ for all $x\in A$, then $|A|\le\aleph_\gamma$

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