{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

5.8ZFC071310 - 1 5.8 ZFC V = L indiscernibles and 01...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
1 5.8. ZFC + V = L, indiscernibles, and Π 0 1 correct arithmetic. We fix M^ = (C,<,0,1,+,-,•, ,log, ω ,c 1 ,c 2 ,...,Y 1 ,Y 2 ,...) as given by Lemma 5.7.30. We work entirely within M^. E.g., we treat C as the universe of points, and regard the elements of the Y k as the internal relations. In particular, if we say that R is an internal relation, then we mean that R is an element of some Y k . If we say that R is an M^ definable relation (first and second order parameters allowed), then we do not necessarily mean that R is an internal relation. However, by Lemma 5.7.30, vii), if R is an M^ definable relation which is bounded , then R is an internal relation; i.e., R is an element of some Y k . In fact, Y k is the set of all bounded M^ definable relations on C. DEFINITION 5.8.1. Functions are always identified with their graphs. We refer to the elements of Y 1 as the internal sets. An important obstacle is that there is no way of showing, in M^, that the family of all internal subsets of an internal set is in any sense internal. E.g., no way of showing that they are all cross sections of some fixed internal binary relation. It would appear that this obstacle is fatal, as it indicates an inability to interpret the power set axiom, despite bounded comprehension, indiscernibility, and infinity. However, in this section, we argue carefully that we can still construct the constructible universe. Because of the explicitness of this construction, we can use indiscernibility to overcome this obstacle within the constructible universe. We first have to develop a pairing function. By an interval, we mean a set [x,y), where x,y C. LEMMA 5.8.1. Let k 1 and F be a k-ary M^ definable function, defined without second order parameters. For all x, {F(y 1 ,...,y k ): y 1 ,...,y k < x} is bounded above. For all x, the restriction of F to [0,x) k is an internal function.
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
2 Proof: Let k,F be as given, and let x C. Let n 1 be such that x and all parameters used in the definition of F are < c n . Let y 1 ,...,y k < x. Let m > n be such that F(y 1 ,...,y k ) < c m . Consider the true statement F(y 1 ,...,y k ) < c m . This is a statement involving c m and certain parameters < c n . By Lemma 5.7.30 ix), F(y 1 ,...,y k ) < c n+1 . The second claim follows immediately by Lemma 5.7.30 vii). QED DEFINITION 5.8.2. For all x C, we write x+1 for the immediate successor of x in <. The above exists by Lemma 5.7.30 v),viii). This is a slight abuse of notation since x+1 already has a meaning, as + is a primitive of M^. However, note that by Lemma 5.7.30 ii),iii), if x < ω then x+1 is also the immediate successor of x in <. LEMMA 5.8.2. Let x,y C, x > 0. There is a unique strictly increasing internal f with dom(f) = [0,x), rng(f) an interval, and f(0) = y. Proof: We first prove a strong form of uniqueness. Suppose x,x’,y C, x,x’ > 0, and let f,g be strictly increasing internal functions, where dom(f) = [0,x), dom(g) = [0,x’), and rng(f),rng(g) are intervals, and f(0) = g(0) = y. Then f,g agree on their common domain. To see this, suppose this is false. By Lemma 5.7.30 viii), let b be < least such that f(b) g(b). Obviously f(b) is the strict sup of the f(c), c < b, and g(b) is the strict sup of the g(c), c < b. Hence f(b) = g(b), which is a contradiction. Hence f,g agree on
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern