See axiom of countable choice in All languages combined, or Wiktionary
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"ref": "2000, Bruno Poizat, translated by Moses Klein, A Course in Model Theory: An Introduction to Contemporary Mathematical Logic, Springer, page 169",
"text": "The finite axiom of choice is not an axiom, but rather a theorem that can be proved from the other axioms. In contrast, there are weak forms of the axiom of choice that are not provable. One example is the axiom of countable choice, which states that if A#x5F;0,A#x5F;1,#x5C;dotsA#x5F;n#x5C;dots form a denumerable set of nonempty sets, their product is nonempty.[…]The axiom of countable choice is constantly used in analysis; it is often hidden so as not to sow confusion in the minds of the students (who are inclined to accept anything desired) or of the professors (who do not like to shake the foundations of the discipline).",
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"ref": "2012, Richard G. Heck, Jr., Reading Frege's Grundgesetze, Oxford University Press, page 271, But, once again, while we can easily prove ∀n[P0n→∃G(∀x(Gx→Fx)∧n=Nx:Gx)]",
"text": "we have no way to infer\n∃R∀n[P0n→∀x(R_nx→Fx)∧n=Nx:R_nx)]\nwithout an axiom of countable choice."
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"text": "2013, Valentin Blot, Colin Riba, On Bar Recursion and Choice in a Classical Setting, Chung-chien Shan (editor), Programming Languages and Systems: 11th International Symposium, APLAS 2013, Proceedings, Springer, LNCS 8301, page 349,\nWe show how Modified Bar-Recursion, a variant of Spector's Bar-Recursion due to Berger and Oliva can be used to realize the Axiom of Countable Choice in Parigot's Lambda-Mu-calculus, a direct-style language for the representation and evaluation of classical proofs."
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