स्नातक (उन्नत)Foundationsनियोजित
The axiom of choice and its equivalents
The axiom of choice, its equivalents Zorn's lemma and the well-ordering theorem, typical applications in algebra and analysis, and its non-constructive consequences.
math.found.axiom-of-choice
इस विषय का पाठ अभी लिखा नहीं गया है। नीचे दी गई रूपरेखा पाठ्यक्रम-मानचित्र का हिस्सा है और इसकी समीक्षा नहीं हुई है।
इस पर आधारित
20 अनिवार्य पूर्वापेक्षाएँ (बिल्कुल नीचे तक)
आप सीखेंगे
- Prove, over ZF, that the axiom of choice, Zorn's lemma and the well-ordering theorem are equivalent.
- Use Zorn's lemma to prove that every vector space has a basis and that every non-zero commutative ring with identity has a maximal ideal.
- Describe results, such as the existence of a non-measurable set of reals, that are provable with choice but not in ZF alone (relative to the consistency of an inaccessible cardinal).
परिभाषा
- Axiom of choice (AC): every family of non-empty sets has a choice function.
- Well-ordering theorem: every set admits a well-order.
- Countable choice and dependent choice (DC) are weaker principles that suffice for much of classical analysis.
प्रमेय
- Over ZF, AC, Zorn's lemma and the well-ordering theorem are equivalent.
- Gödel: if ZF is consistent, so is ZFC. Cohen (1963): if ZF is consistent, so is ZF + ¬AC.
- AC implies that some subset of ℝ is not Lebesgue measurable (Vitali). Solovay showed that, if an inaccessible cardinal is consistent with ZFC, then ZF + DC is consistent with every set of reals being Lebesgue measurable.
आम भ्रांतियाँ
- Thinking choice is only needed for uncountable families; choosing from countably many two-element sets can already require it.
- Believing the Banach–Tarski paradox refutes AC; it shows that some sets cannot be assigned a volume invariant under rigid motions.
- Assuming every proof that uses Zorn's lemma can easily be made constructive.
आगे ले जाता है
वैकल्पिक खोजें
- Measure theory and integration Vitali's construction of a non-measurable set uses AC; this is one reason Lebesgue measure is defined on a σ-algebra of sets rather than on all subsets of ℝ.
- Functional analysis The Hahn–Banach theorem is usually proved with Zorn's lemma, although it follows from the strictly weaker ultrafilter lemma.
- Advanced linear algebra Over ZF, 'every vector space has a basis' is equivalent to the axiom of choice (Blass, 1984).
आगे पढ़ने के लिए
- Thomas Jech, The Axiom of Choice (1973)
- Kenneth Kunen, Set Theory: An Introduction to Independence Proofs (1980)