query: Can existence of aleph one be proved without the power set axiom?
positive_passages.docid: 19924
positive_passages.text: Uncountable ordinals without power set axiom
negative_passages.docid: 142267
negative_passages.text: If two sets have the same cardinality, then so do their power sets. Converse can't be answered?
negative_passages.docid: 93771
negative_passages.text: Prove that for any set $A$ there is some $x \notin A$
negative_passages.docid: 156161
negative_passages.text: Banach-Tarski theorem without axiom of choice
negative_passages.docid: 122324
negative_passages.text: Is there a proof of Gödel's Incompleteness Theorem without self-referential statements?
negative_passages.docid: 28223
negative_passages.text: Prove/Disprove that if two sets have the same power set then they are the same set
negative_passages.docid: 139992
negative_passages.text: Is power-associativity finitely axiomatizable?
negative_passages.docid: 121847
negative_passages.text: Don't we need the axiom of choice to choose from a non-empty set?
negative_passages.docid: 4782
negative_passages.text: Can we distinguish $\aleph_0$ from $\aleph_1$ in Nature?
negative_passages.docid: 21782
negative_passages.text: Is $\aleph_0 = \mathbb{N}$?
negative_passages.docid: 31389
negative_passages.text: consistency of ZF\{Power Set axiom}
negative_passages.docid: 110957
negative_passages.text: No non-trivial clopen sets in $\mathbb{R}$? How to give a direct proof?
negative_passages.docid: 16534
negative_passages.text: Can $(\Bbb N,\leq)$ have an $\aleph_0$-categorical theory (in a larger language)?
negative_passages.docid: 104744
negative_passages.text: Can sets of cardinality $\aleph_1$ have nonzero measure?
negative_passages.docid: 129813
negative_passages.text: Is there an easy proof for ${\aleph_\omega} ^ {\aleph_1} = {2}^{\aleph_1}\cdot{\aleph_\omega}^{\aleph_0}$?
negative_passages.docid: 80019
negative_passages.text: $\varepsilon$-number countability without choice
negative_passages.docid: 32009
negative_passages.text: Prove $\;\big((p\rightarrow q) \lor (p \rightarrow r)\big) \rightarrow (q\lor r)\equiv p \lor q \lor r$ without use of a truth table.