query: Without Foundation, can axiom of choice be derived from $\forall\alpha\in\mathbf{Ord} (P(\alpha)\text{ can be well ordered})$?
positive_passages.docid: 42314
positive_passages.text: Without appealing to choice, can we prove that if $X$ is well-orderable, then so too is $2^X$?
negative_passages.docid: 153935
negative_passages.text: A weaker Axiom of Infinity?
negative_passages.docid: 30587
negative_passages.text: Uncountable well ordered set in $Z^-$ theory.
negative_passages.docid: 17624
negative_passages.text: Axiom of Choice iff Every set has a choice function
negative_passages.docid: 152272
negative_passages.text: Can you explain the "Axiom of choice" in simple terms?
negative_passages.docid: 58447
negative_passages.text: Is axiom of choice required for there to be an infinite linearly independent set in a (non-finite-dimensional) vector space?
negative_passages.docid: 126569
negative_passages.text: Can we find a well ordering on an infinite set with no largest element?
negative_passages.docid: 114550
negative_passages.text: Well-ordering theorem and second-order logic
negative_passages.docid: 98580
negative_passages.text: Lebesgue-measurable sets requiring the Axiom of Choice to construct
negative_passages.docid: 35972
negative_passages.text: Well ordering principle for rationals
negative_passages.docid: 50506
negative_passages.text: What concept does an open set axiomatise?
negative_passages.docid: 52451
negative_passages.text: Ordered Triples and Set Theory
negative_passages.docid: 70262
negative_passages.text: Can the axiom of choice be used to define the real numbers?
negative_passages.docid: 153634
negative_passages.text: A statment about the equivalence between the Axiom of Choice and the surjectivity of the projections.
negative_passages.docid: 19924
negative_passages.text: Uncountable ordinals without power set axiom
negative_passages.docid: 69337
negative_passages.text: Zorn's Lemma And Axiom of Choice
negative_passages.docid: 3902
negative_passages.text: Does $k+\aleph_0=\mathfrak{c}$ imply $k=\mathfrak{c}$ without the Axiom of Choice?