5.7Transf071310 - 1 5.7 Transfinite induction comprehension...

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1 5.7. Transfinite induction, comprehension, indiscernibles, infinity, Π 0 1 correctness. We now fix M# = (D,<, ,NAT,0,1,+,-,•, ,log,d 1 ,d 2 ,...) as given by Lemma 5.6.18. While working in M#, we must be cautious. a. The linear ordering < may not be internally well ordered. In fact, there may not even be a < minimal element above the initial segment given by NAT. b. We may not have extensionality. Note that we have lost the internally second order nature of M* as we passed from M* to the present M# in section 5.6. The goal of this section is to recover this internally second order aspect, and gain internal well foundedness of <. To avoid confusion, we use the three symbols =, , . Here = is the standard identity relation we have been using throughout the book. DEFINITION 5.7.1. We use for extensionality equality in the form x y ( z)(z x z y). DEFINITION 5.7.2. We use as a special symbol in certain contexts. DEFINITION 5.7.3. We write x if and only if x has no elements. We avoid using {x 1 ,...,x k } out of context, as there may be more than one set represented in this way. DEFINITION 5.7.4. Let k 1. We write x {y 1 ,...,y k } if and only if ( z)(z x (z = y 1 ... z = y k )). LEMMA 5.7.1. Let k 1. For all y 1 ,...,y k there exists x {y 1 ,...,y k }. Here x is unique up to . Proof: Let y = max(y 1 ,...,y k ). By Lemma 5.6.18 iv),
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2 ( x)( z)(z x (z y (z = y 1 ... z = y k ))). The last claim is obvious. QED DEFINITION 5.7.5. We write x <y,z> if and only if there exists a,b such that i) x {a,b}; ii) a {y}; iii) b {y,z}. LEMMA 5.7.2. If x <y,z> w x, then w {y} w {y,z}. If x <y,z> x <u,v>, then y = u z = v. For all y,z, there exists x <y,z>. Proof: For the first claim, let x,y,z,w be as given. Let a,b be such that x {a,b}, a {y}, b {y,z}. Then w = a w = b. Hence w {y} w {y,z}. For the second claim, let x <y,z>, x <u,v>. Let x {a,b}, a {y}, b {y,z} x {c,d}, c {u}, d {u,v}. Then (a = c a = d) (b = c b = d) (c = a c = b) (d = a d = b). Since a = c a = d, we have y = u (y = u = v). Hence y = u. We have b {y,z}, d {y,v}. If b = d then z = v. So we can assume b d. Hence b = c, d = a. Therefore u = y = z, y = u = v. For the third claim, let y,z. By Lemma 5.7.1, let a {y} and b {y,z}. Let x {a,b}. Then x <y,z>. QED DEFINITION 5.7.6. Let k 2. We inductively define x <y 1 ,...,y k > as follows. x <y 1 ,...,y k+1 > if and only if ( z)(x <z,y 3 ,...,y k+1 > z <y 1 ,y 2 >). In addition, we define x <y> if and only if x = y. LEMMA 5.7.3. Let k
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This note was uploaded on 08/05/2011 for the course MATH 366 taught by Professor Joshua during the Fall '08 term at Ohio State.

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5.7Transf071310 - 1 5.7 Transfinite induction comprehension...

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