Question

Consequences of a class bijection $V\times V\to V$

score 61 · Answer 1 · 2020-12-02 02:49:08

61

Views

Consequences of a class bijection $V\times V\to V$

Published on 02 Dec 2020 - 2:49

score 573 · Answer 2 · 2020-12-02 15:49:35

573

Views

Basis of $\mathbb R$ seen as a $\mathbb Q$-vector space

Published on 02 Dec 2020 - 15:49

score 181 · Answer 3 · 2026-03-25 23:35:07

181

Views

Constructively embedding $\mathbb{Q}^\mathbb{N}$ into $\mathbb{R}$

Published on 25 Mar 2026 - 23:35

score 106 · Answer 4 · 2020-12-05 16:51:58

106

Views

How is axiom of choice utilized within the given proof?

Published on 05 Dec 2020 - 16:51

score 83 · Answer 5 · 2020-12-14 13:34:31

83

Views

Equivalent statement of AC or not?

Published on 14 Dec 2020 - 13:34

score 43 · Answer 6 · 2020-12-15 01:01:27

43

Views

Is this partition axiom equivalent to axiom of choice over ZF?

Published on 15 Dec 2020 - 1:01

score 121 · Answer 7 · 2026-03-29 17:50:04

121

Views

Is it true ZF can't prove there is a choice function on the Turing degrees?

Published on 29 Mar 2026 - 17:50

score 402 · Answer 8 · 2020-12-19 14:33:51

402

Views

Does the law of excluded middle imply the axiom of choice?

Published on 19 Dec 2020 - 14:33

score 25 · Answer 9 · 2020-12-19 19:49:37

25

Views

equivalent versions of axiom of choice

Published on 19 Dec 2020 - 19:49

score 122 · Answer 10 · 2026-03-27 22:03:53

122

Views

Do I understand this proof for AoC $\implies$ well-ordering theorem correctly?

Published on 27 Mar 2026 - 22:03

score 29 · Answer 11 · 2026-04-06 07:08:51

29

Views

Is this proposition regarding equipotence true, without the needing axiom of choice?

Published on 06 Apr 2026 - 7:08

score 59 · Answer 12 · 2020-12-28 11:41:34

59

Views

Are these two definitions of finiteness equivalent?

Published on 28 Dec 2020 - 11:41

score 352 · Answer 13 · 2020-12-28 21:17:59

352

Views

Can you "construct" a category from another category?

Published on 28 Dec 2020 - 21:17

score 198 · Answer 14 · 2020-12-30 21:20:05

198

Views

Proof that non-well-orderable sets are not equinumerous to their own cardinals (using Scott's trick)

Published on 30 Dec 2020 - 21:20

score 234 · Answer 15 · 2026-04-03 07:13:13

234

Views

The Axiom of Choice implies that every topological vector space has a balanced base and every locally convex space has a balanced convex local base.

Published on 03 Apr 2026 - 7:13

List Question

Consequences of a class bijection $V\times V\to V$

Basis of $\mathbb R$ seen as a $\mathbb Q$-vector space

Constructively embedding $\mathbb{Q}^\mathbb{N}$ into $\mathbb{R}$

How is axiom of choice utilized within the given proof?

Equivalent statement of AC or not?

Is this partition axiom equivalent to axiom of choice over ZF?

Is it true ZF can't prove there is a choice function on the Turing degrees?

Does the law of excluded middle imply the axiom of choice?

equivalent versions of axiom of choice

Do I understand this proof for AoC $\implies$ well-ordering theorem correctly?

Is this proposition regarding equipotence true, without the needing axiom of choice?

Are these two definitions of finiteness equivalent?

Can you "construct" a category from another category?

Proof that non-well-orderable sets are not equinumerous to their own cardinals (using Scott's trick)

The Axiom of Choice implies that every topological vector space has a balanced base and every locally convex space has a balanced convex local base.

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