Question

Classification of connected one dimensional manifolds and Zorn lemma

score 73 · Answer 1 · 2026-03-29 10:18:07

73

Views

Classification of connected one dimensional manifolds and Zorn lemma

Published on 29 Mar 2026 - 10:18

score 125 · Answer 2 · 2019-11-18 12:13:30

125

Views

Only base in definition of compactness - do we need AC?

Published on 18 Nov 2019 - 12:13

score 216 · Answer 3 · 2019-11-18 13:00:03

216

Views

"Indexed" version of compactness and Axiom of Choice

Published on 18 Nov 2019 - 13:00

score 190 · Answer 4 · 2019-11-23 20:37:04

190

Views

Is the axiom of choice needed in proving that metric spaces in which every infinite subset has a limit point are compact?

Published on 23 Nov 2019 - 20:37

score 45 · Answer 5 · 2019-11-26 02:01:03

45

Views

First countable + sequentially closed implies closed without AC

Published on 26 Nov 2019 - 2:01

score 68 · Answer 6 · 2019-11-26 02:29:27

68

Views

Given sets $A,B$, there exists injection $f:A\to B$ or injection $g:B\to A$

Published on 26 Nov 2019 - 2:29

score 93 · Answer 7 · 2019-11-28 19:12:23

93

Views

In the basic Fraenkel (permutation) model, any functions from $\omega$ to $P(A)$ has finite range.

Published on 28 Nov 2019 - 19:12

score 98 · Answer 8 · 2019-12-02 09:26:27

98

Views

Are sets in ZFC that aren't in certain models of ZF without AC included as classes or some other Platonic concept?

Published on 02 Dec 2019 - 9:26

score 53 · Answer 9 · 2019-12-03 09:21:46

53

Views

Within the second Fraenkel model, show that any function from $A\rightarrow A$ has the following property

Published on 03 Dec 2019 - 9:21

score 87 · Answer 10 · 2019-12-09 01:29:57

87

Views

Axiom of Choice --Example Problem

Published on 09 Dec 2019 - 1:29

score 116 · Answer 11 · 2026-03-26 04:24:52

116

Views

Constructive mathematics plus existence of discontinuous functions

Published on 26 Mar 2026 - 4:24

score 63 · Answer 12 · 2019-12-11 23:58:27

63

Views

show that there is a set $ A_0 \subseteq A $ such that $ card(B) = card(A_0 ) \le card(A) $

Published on 11 Dec 2019 - 23:58

score 130 · Answer 13 · 2026-03-25 22:30:02

130

Views

Prove that a countable inductive poset is directed complete.

Published on 25 Mar 2026 - 22:30

score 190 · Answer 14 · 2019-12-14 13:35:35

190

Views

Is it possible to avoid the axiom of choice?

Published on 14 Dec 2019 - 13:35

score 185 · Answer 15 · 2019-12-16 01:39:45

For Infinite $B\subseteq A$, do $|A\times A|=|A|$, $B\subset A$, and $|B|<|A|$ Imply $|A\setminus B|=|A|$ without the axiom of choice?

List Question

Classification of connected one dimensional manifolds and Zorn lemma

Only base in definition of compactness - do we need AC?

"Indexed" version of compactness and Axiom of Choice

Is the axiom of choice needed in proving that metric spaces in which every infinite subset has a limit point are compact?

First countable + sequentially closed implies closed without AC

Given sets $A,B$, there exists injection $f:A\to B$ or injection $g:B\to A$

In the basic Fraenkel (permutation) model, any functions from $\omega$ to $P(A)$ has finite range.

Are sets in ZFC that aren't in certain models of ZF without AC included as classes or some other Platonic concept?

Within the second Fraenkel model, show that any function from $A\rightarrow A$ has the following property

Axiom of Choice --Example Problem

Constructive mathematics plus existence of discontinuous functions

show that there is a set $ A_0 \subseteq A $ such that $ card(B) = card(A_0 ) \le card(A) $

Prove that a countable inductive poset is directed complete.

Is it possible to avoid the axiom of choice?

For Infinite $B\subseteq A$, do $|A\times A|=|A|$, $B\subset A$, and $|B|<|A|$ Imply $|A\setminus B|=|A|$ without the axiom of choice?

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