By Bela Sz. -Nagy

**Read Online or Download Appendix to Frigyes Riesz and Bela Sz. -Nagy Functional Analysis... PDF**

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**Extra resources for Appendix to Frigyes Riesz and Bela Sz. -Nagy Functional Analysis... **

**Example text**

3) T is consistent and complete. In this case any sentence is exactly in one of T, F hence N contains no sentence. In case 3 note that we may define truth/falsehood of sentences syntactically as follows: a sentence P is true if it is in T ; and it is false if it is in F . This is a metadefinition and implies that a sentence is either true or false and cannot be both true and false. However one can prove (cf. G¨odel) a sentence in set theory whose translation into English says that if set theory is consistent then it is incomplete.

2. ) 3. ) ... 635. ). Here is an example of a theorem and a proof. It is a theorem in any theory. It is referred to as the quantified modus ponens. 2. ((∃xP (x)) ∧ (∀x(P (x) → Q(x)))) → (∃xQ(x)). Proof. 1. (∃xP (x)) → P (cP ) (quantifier axiom). 2. (∀x(P (x) → Q(x))) → (P (cP ) → Q(cP )) (quantifier axiom). 3. ((∃xP (x)) ∧ (∀x(P (x) → Q(x)))) → (P (cP ) ∧ (P (cP ) → Q(cP ))) (by 1, 2). 4. ((P (cP ) ∧ (P (cP ) → Q(cP )))) → Q(cP ) (tautology: modus ponens). 5. Q(cP ) → ∃xQ(x) (quantifier axiom).

33. A theory T is maximal if it is consistent and if any consistent theory containing T coincides with T . 34. Metaprove (using the above) that a consistent theory is complete if and only if it is maximal. 35. One can ask if any consistent theory is contained in a complete theory. The latter question is a metasentence that involves quantifiers so it cannot be metaproved. The mirror of this in mathematical logic is a theorem (a consequence of what is called Zorn’s lemma). 36. The moral of the last few exercises is that the finitistic requirement for metaproofs is such that very few metasentences of interest can be actually metaproved.