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0 votes
3 answers
71 views

Construction of Proof: Zorn's lemma implies Axiom of choice

I have come across the prove that [Zorn's Lemma ==> AC] but am confused about the central statement, namely that we can take a set of all choice functions on subsets of X (lets just call it X, I ...
CopperCableIsolator's user avatar
13 votes
3 answers
1k views

Is the Axiom of Choice inconsistent with Countable Additivity?

Consider a fair lottery among a countably infinite number of people. The Axiom of Countable Additivity says this is impossible to construct: If all people have a positive (and equal) probability of ...
Joseph Camacho's user avatar
3 votes
1 answer
84 views

Do you need axiom of choice to proof $2^\mathbb{N}$ is nonempty? [duplicate]

Define $2^\mathbb{N}$ as the set $$ 2^\mathbb{N} = \prod_{i \in \mathbb{N}} \{0,1\} = \{ f: \mathbb{N} \to \{0,1\} \}.$$ It seems we do not need the axiom of choice to show that this set is non-empty; ...
Inzinity's user avatar
  • 1,773
2 votes
1 answer
82 views

Can cardinality be defined in ZF-Regularity without the axiom of choice and without Scott's trick?

In ZF-Regularity, can we define a notion of cardinality such that for all sets $A$ and $B$, $card(A) = card(B)$ iff there exists a bijection from $A$ to $B$? If we add a function symbol C to the ...
Hussein Aiman's user avatar
0 votes
0 answers
33 views

A countable union of countable sets is countable requires the axiom of choice [duplicate]

By countable sets, I refer to countably infinite sets. We can very well prove that the natrual numbers and $\mathbb{N} \times \mathbb{N}$ have the same cardinality, for a bijection take the Cantor ...
Shthephathord23's user avatar
2 votes
1 answer
59 views

Relation between Axiom of Foundation and $\in$-induction

At the end of an intro to set theory course, we were introduced the Axiom of Foundation and the Principle of $\in$-induction as one of its consequences. I found it easy to prove that, assuming the ...
carfog's user avatar
  • 322
2 votes
1 answer
98 views

$AC_\omega$ is weaker than $AC$, though not provable in ZF without $AC$?

According to the wikipedia article on the Axiom of countable choice, The axiom of countable choice ($AC_ω$) is strictly weaker than the axiom of dependent choice (DC), (Jech 1973) which in turn is ...
Hank Igoe's user avatar
  • 1,416
1 vote
0 answers
70 views

Can this choice-collection principle prove the well ordering theorem without power set axiom?

Choice Collection: if $\phi(x,y)$ is a formula in which "$B$" doesn't occur free, and "$x,y$" are among its free variables; then: $$\forall A \exists B \forall x \in A \, (\exists ...
Zuhair's user avatar
  • 4,631
1 vote
0 answers
125 views

Is the Axiom of choice intuitive? How was it first introduced?

I will refer to the Axiom of choice as ($AC$). As far as I know, the widely accepted axioms of set theory is the Zermelo-Fraenkel axioms together with the $AC$. But the last axiom seems to be the most ...
Tereza Tizkova's user avatar
3 votes
4 answers
251 views

Connection between the many different definitions of the Axiom of Choice

I'm currently trying to write a paper on the Axiom of Choice. With my research I have found one very simple definition of the Axiom of Choice : "Let X be a non-empty set of non-empty sets. There ...
Robin McManus's user avatar
2 votes
0 answers
86 views

How much choice is preserved in Gitik models of ZF?

In a Gitik model of $\sf ZF$ choice is obviousely lost since all uncountable cardinals are singular and of cofinality $\omega$. Now that choice is lost doesn't by itself entail that weaker forms of ...
Zuhair's user avatar
  • 4,631
2 votes
1 answer
80 views

Adding the "Step-by-step injective extensions" axiom to ZF

Update What I was trying to do here was to come up with an axiom that could replace the axiom of countable choice. Although the effort went down in flames, I now more strongly appreciate and respect ...
CopyPasteIt's user avatar
  • 11.5k
0 votes
1 answer
73 views

Do tree choice lead to set choice?

In a tree that is mono-rooted, lets call the root node as the primary node, the nodes connected to it by edges as the secondary nodes, and those connected to those by edges as the tertiary nodes, etc.....
Zuhair's user avatar
  • 4,631
3 votes
1 answer
81 views

Is the existence of arbitrary definable partial functions bounded by a set equivalent to the axiom of choice?

Is the axiom of choice (125) equivalent to the following axiom (120)? If we take (120) and the companion axioms (180) and (190), can we demote specification (140) and replacement (160) to theorems? ...
Greg Nisbet's user avatar
  • 11.9k
0 votes
0 answers
57 views

uses and proofs of the cardinality of the continuum

after reading this https://en.wikipedia.org/wiki/Cardinality_of_the_continuum I wonder what good is it as a tool in mathematics if there is no proof. is it merely useful to prove that it is ...
acacia's user avatar
  • 249

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