3   Forcing  Definition   Forcing                  is called a forcing poset  or forcing  if it is a partial order and        and    is the largest element  Elements of   are called condition   p   q is interpreted as   p  is stronger than q    q is weaker than p   Note   Unconventional    Alternative   Jerusalem convention   Interpret p   q as  q is stronger than p    We are not  following the Jerusalem convention   Definition   Compatible   p and q are compatible  if there is r   p   q  Otherwise  we say they are incompatible   which we write as p   q    Definition   Antichain   A   P is an antichain  if any two distinct elements of A are incompatible   Definition   Dense   D     is dense  if   p   T O D O   Definition   Filter   F      is called a filter  if   a    p   q   F     r   F   r   p   q   b    p   F     q       q   p   q   F  If F only has property  a   we call it a filter base  and then        p         q   F   q   p       is the filter generated from  F   Note   Filters cannot  contain incompatible elements    Definition   D generic   If D is a family of dense sets  then F is called D generic  if F is a filter and   D   D   D   F       Theorem 3 1  Assuming that   D is countable  Then  there is a D generic  filter   Proof  Let D     D n   n        Pick p 0   D 0 arbitrarily and define by recursion p i   1 by picking some q   p i with q   D i   1   Then   p i   i       is a filter base  so let F be the filter generated by it  Then it is D generic  by construction     Main example  Fix any sets X and Y     Fn   X   Y         p   p is a finite fucntion with dom   p     X and range   p     Y         Define p   q   p   q and            What does p   q mean       p   q     x   X   x   dom   p     dom   q     p   x     q   x         Lemma 3 2  Assuming that   F     is a filter  Then    F is a function   Lemma 3 3  D x       p       x   dom   p     is a dense set  If D       D x   x   X    and F is D generic  then dom     F     X   Proof  If x   X  find p   F   D x  then x   dom   p    TODO    3 1   Cohen Forcing  Example 1        Fn       2     If F is D generic  then   F       2  by above    Lemma 3 4  Assuming that   Fix f       2 and define      N f       p         u   p   u     f   u           F is D     N f    Then    F   f   Corollary 3 5  There is no F that is D     N f   f       2   generic   However   if M is a countable transitive model and you consider      N f       N f   f   M         then by Theorem 3 1  there is a D   N M generic  And for this F  TODO  Example 2  Fn   X   Y         as before      R y       p       y   range   p     R       R y   y   Y       Lemma 3 6  Assuming that   F is D   R generic  Then    F   X   Y is a surjection   Corollary 3 7  Assuming that     Y       X    Then  there is no D   R generic   However   if M is a countable transitive model and  e g  TODO  Example 3  Fn   X   Y   2    Assume X is infinite  Consider      E y   y         p         x   X   p   x   y     p   x   y             This is dense for y   y        E       E y   y     y   y     Y         Lemma 3 8  Assuming that   F is D   E generic  Then  there is an injection from Y into P   X     Proof  Fix y and define      A y       x   X       F     x   y     1         E y   y   guarantees that y   A y is an injection     Corollary 3 9  Assuming that     Y       P          Then  there is no D   E generic   However   if M is a countable transitive model of ZFC   CH    is countable and so a D   E generic  exists   Recap  M a countable transitive model of ZFC  or a sufficiently large finite fragment        M  dense   filter    D generic   Example  Fn       2   produces a new function f       2  Cohen forcing    Example  Fn   X   Y   produces a surjection f   X   Y  Fn       Y   COLAPSE of Y   Example  Fn   X   Y   2   produces  an injection f   Y   P   X     If M is a countable transitive model of ZFC  then D M       D   P dense   D   M   is countable  so by Theorem  we have a D M generic   Definition 3 10   P generic over M    We say F is   generic  over  M if it is D M generic  These always exist if M is a countable transitive model   GOAL  Build an extension M   G   such that M   M   G    M   G   a countable transitive model of ZFC  G   M   G   and M   G   is minimal   Names  Idea   Think of elements of   as  truth values  for the von Neumann construction      Name 0         Name     1                 Name         Name                 Name         Then      Name             Ord Name         is the proper class of all names   Consider        0   1    Then Name     V   Since this is a recursive definition using absolute concepts  being a name is absolute for transitive models            M     is a    name     Name     M       TODO  Examples      Name 1          p             p       Name 2        The name for a set that contains   with value p         p q           p   q            The name for whatever   p describes with value q    Interpretation  If F      we interpret a   name as follows       val       F         val       F       p   F         p               Important   This is a recursive definition   Thus  the valuation is absolute for transitive models containing   and F   Example   1       clearly val       F          2      p      val     p   F               p   F   p   F      3      p q      val     p q   F           q   F           q   F   p   F       q   F   p   F     The relationship between p and q affects these possibilities   Example  if q   p and F is a filter  then       is impossible   Example  if  q     and F is a non empty filter  then   is impossible   Example  if p   q and F is a filter  then           is impossible   Definition 3 11   Generic extension    The  generic  extension  for any countable transitive model  M and any F     where     M is     M   F         val       F         Name     M         Obviously  M   F   is a countable set with     M   F    Example  1    Also  by definition  M   F   is transitive   Note      M   F     Extensionality   Foundation       Need to show   1    M   M   F      2    F   M   F      3    M   F     ZFC    4    M   F   is minimal   Canonical Names  Definition 3 12   Canonical name    Let x   M  Define by recursion the canonical name for  x by         w           y           y   x         Lemma 3 13  val   x     F     x if      F   Proof  Induction     Corollary 3 14  M   M   F   if      F   Alternative construction of canonical names without    is on Example Sheet 2                 p     p     p             Lemma 3 15  val       F     F   Proof  Calculate      val       F       val   p     F     p   F       p   p   F    by previous lemma    F        Corollary 3 16  F   M   F     Remark   If N is a countable transitive model with M   N and F   N  then M   F     N   by absolute ness of val       F      Warm up  Suppose         Name    Define          up                                               Pairing   Then      val   up             F         val       F     val       F             up  stands for unordered pair    Corollary 3 17  M   F     Pairing  if      F    Union   If   is a name  define      u                 r           p   q   s t        p                 q         r   p   q         Claim  val   u     F       val       F   if F is a filter   Proof  Suppose z   val   u     F     So z   val         F   for some         r     u   with r   F   So       p   q with       p                q        r   p   q   So p   q   F  Hence val       F     val       F    z   val         F     val       F     So z     val       F     Conversely  suppose z     val       F    Then   y   z   y   val       F     z           q       with q   F  y                 with p   F    Hence since F a filter  find r   p   q with r   F   Then         r     u    so z   val   u     F       TODO  Further Recap  We proved      M   G     Extensionality   Foundation   Pairing   Union       Homework was  Think about why power set is not easy    Union  proof was   collect all natural candidates of names for elements and assign the natural values   Problem  If you try to do this for power set  neither the  natural candidates for names  nor the  natural values  are obvious  It ll turn out that they are  obvious in the end  but that requires some assistance   Note  Separation and Replacement are even worse   One remaining easy axiom  AC  By well ordering theorem  AC holds if and only if   x           i   i   x     injection  If x   M   F    then there is     Name such that x   val       F     Notation  I write dom                   p         p           Consider dom        In M  I have i   dom           for some ordinal    In M   F    define      y   min   i         val       F     y       dom               Call this i     Then i     x     is an injection  So  AC holds in M   F     Forcing  Definition   Forcing language   Fix M a countable transitive model of ZFC      M a forcing poset   We call the language      L forcing     L              names       the forcing language  over  M    Definition   Interpretation of forcing language   If G is a   generic  over M and   is in the forcing language  we say      M   G           if and only if      M   G     v         where v           val       G     Definition   Semantic forcing predicate   p   M        Forall G   generic  over M with p   G  M   G          p forces     We often omit M       Two theorems at the heart of forcing    1    Forcing Theorem   If G is a   generic  over M  then      M   G           p   G   p            2    Definability Theorem     p      is absolute ly definable for transitive models containing M   We are going to do the following    a  Define the syntactic forcing predicate      that is absolute ly definable    b  Prove the Forcing Theorem for        c  Derive that            Lectures 11 and 12   Observations  under the assumption of the Forcing Theorem     1    If q   p and  p      then  q        2    If  p     x      then there is a name   and q   p such that  q             if  p     x       x   and p   G  then by definition M   G       x       x    so there is a name   such that M   G              By Forcing Theorem  find r   G such that  r            Find q   p   r   q   G  By  1    q               Proof of Separation in  M   G     Let     x   x 1       x n   be an L   formula  not in the forcing language   and x   M   G    Need a name for      A       z   x   M   G         z   p              For readability  we now drop parameters p      Fix   such that val       G     x  Define                    p         dom         p                           By the Definability Theorem    is a name in M   Claim  val       G     A         If z   val       G    then there is       p       such that z   val       G    p   G  Since       p        we get     dom         p                    Together with p   G  we get      M   G                       z   x       z         Hence z   A         If z   A  then z   x and M   G         z    So there is     dom        z   val       G    By the Forcing THeorem    p   G   p             Also  there is q   G and  q          Find r   G  r   p   q such that  r       r             Hence       r        so z   val       G     val       G       Proof of power set   Example Sheet 3     Proof of Replacement   Since we already have Separation  it s enough to show the following   if   is a functional formula and x   M   G    then there is R   M   G   such that      M   G       y   x     z   R       y   z   p              as before  we suppress parameters for notational ease    We work in M and identify a name   for R  Fix   such that x   val       G    Find   large enough such that dom         V    Consider      p                 p                p        a name   By Definability Theorem  this is an L   formula   By Levy Reflection Theorem  find       such that   is absolute between V   and V   M  Define                    1         V         and R     val       G     Now we verify      holds  Fix y   x  y   val       G    Since   is functional  we know that             holds in M   G   for some    By Forcing Theorem  there is p   G such that  p                So M       p         By absolute ness  V         p        This means       V   such that  p                 Thus  val       G     val       G     R       Definition   Dense below p   D     is called dense below  p if   q   p     r   q   r   D   Lemma 3 18  Assuming that   G is   generic  over M  E      E   M  Then    i  If E is dense below p  q   p  then E is dense below q    ii  If   r   E is dense below r   is dense below p  then E is dense below p   iii  Either G   E     or   q   G    r   E   r   q    iv  If p   G  E is dense below p  then G   E       Proof  Example Sheet 3    Definition of the syntactic forcing relation  p                 Two recursions   First          by recursion on Name     Then          without recursion   Then the rest by recursion on formula complexity   p       0     1  if and only if           0   s 0       0   q   p   q   s 0         1   s 1       1     q   s 1   q       0     1     is dense below p     and           1   s 1       1   q   p   q   s 1         0   s 0       0     q   s 0   q       0     1     is dense below p     Remark   This is a recursion on Name     p       0     1  if and only if        q   p           s       1     q   s   q           0         is dense below p   Remark   No recursion involved  just            Recursion on complexity of formulas    p            if and only if  p       and  p         p          if and only if   q   p   q           p       x       x    if and only if    r         r                 Remark   These definitions remind us of Kripke semantics for intuitionistic logic    Lemma 3 19  The following are equivalent    i  p          ii    r   p   r          iii    r   r         is dense below     Proof  If this is true for   of the form   0     1 and of the form   0     1  then it s true for all formulas   For   atomic  we get that  ii     i  and  ii     iii  are trivial    i     ii  follows from Lemma 3 18  i     iii     i  follows from Lemma 3 18  ii      Strategy   Theorem 3 20   Syntactic Forcing Theorem    M   G        if and only if    p   G   M   r         Corollary 3 21   Definability Theorem    p      if and only if  p            p   M     M   p         Corollary 3 22   Forcing Theorem    M   G        if and only if    p   G   p       This corollary is immediate from combining the two previously stated results   Proof of Definability Theorem from Syntactic Forcing Theorem      This is just Syntactic Forcing Theorem      Suppose  p      By Lemma 3 18  we need to show        r   p   r             is dense below p   Suppose not  Then I find q   p such that   r   q   r          So  q           If q   G  then p   G  Since  p      M   G        Since  q          M   G          Contradiction     Proof of the Syntactic Forcing Theorem  The Theorem for   can be proved by induction on Name     The Theorem for   can be proved once the Theorem has been established for     Rest is induction on formula complexity   Let s do   and leave the rest to Example Sheet 3  Assume we have proved it for    We will prove it for           M   G          Consider      D       p   p       or p                 By definition of         this is dense  Find p   G   D   By assumption   p          so  p               Let p   G such that  p          By definition    q   p   q          If M   G            thus M   G        By induction hypothesis  find r   G   r        So q   r   p   q   G  By Lemma   q        contradiction to    q   p   q              Note   The Forcing Theorem is very useful  The inner details of the proof are not very important though  We won t really discuss these details once we have proved the theorem    Continuing the proof of Syntactic Forcing Theorem  Assume Syntactic Forcing Theorem for     and prove if for       Want to show   p       0     1 if and only if        q           s       1     q   s   q           0         is dense below p   Proof      Assume that M   G       0     1  i e  val     0   G     val     1   G    Thus  there is       s       1 such that val       G     val     0   G      M   G           0  and s   G  So by Syntactic Forcing Theorem for    find r   G such that  r           0  Find p   r   s such that p   G   Claim that D is dense below p  Let q   D and pick the above       s    Then we have q   p   s and  q           0  since q   p   r        Assume p   G   p       0     1  i e  D is dense below p  By Lemma 3 18  iv   find q   G   D  thus there is       s       1 such that q   s   q           0  Since q   G  s   G  we have val       G     val     1   G    By Syntactic Forcing Theorem for      we have val       G     val     0   G    So val     0   G     val     1   G       Now we prove it for       We prove this by induction on name rank  so assume that Syntactic Forcing Theorem for     is true for names   0     1 with smaller name rank   Reminder  we need to show M   G       0     1 if and only if   p   G   p       0     1   TODO  double check the below  Proof      Assume p   G such that  p       0     1  Need to prove val     0   G     val     1   G    We ll show that the fact that D 0 is dense below p implies val     0   G     val     1   G    The other direction is the same proof but with 0 and 1 flipped   Proof of this  Let x   val     0   G    so find     0   s 0       0 such that s 0   G and x   val     0   G    So  the corresponding D 0 is dense below p  Find q   s 0   p such that q   G  Then D 0 is dense below q  So find r   q  such that r   G   D 0   Note that r   q   s 0  so r   s 0    Since r   D 0 and r   s 0  find     1   s 1       1 such that  r   s 1   r       0     1  By induction hypothesis  r   G and  r       0     1  which implies x   val       v     val     1   G    since     1   s 1       1 and s 1   G        Assume val     0   G     val     1   G    Consider the set     D       r   r       0     1 or   r 0       0   s 0       0   r   s 0         1   s 1       1     q     q   s 1   q       0     1     q   r     r 1       1   s 1       1   r   s 1         0   s 0       0     q     q   s 0   q       0     1     q   r         Claim  D is dense  If  p      then p   D  so nothing to show  If  p         0     1  then there is some D 0   D 1 that fails to be dense below p   We ll show  if D 0 is not dense below p  then we can find r   p such that   r 0 holds    Other proof  D 1 not dense below p     r 1 is flipping 0 and 1    Suppose  p         0     1 and there is     0   s 0       0 such that D 0 is not dense below p  This means  there is r   p such that        q   r   q   s 0         1   s 1       1     q   s 1   q       0     1         So  we get      r   s 0         1   s 1       1     for any q that satisfies  q   s 1   q       0     1  It can t be compatible with r  finishes the proof of the claim that D is dense     Summary  We now have that D is dense       D     r   r       0     1 or   r 0 or   1 r         Claim 2  If r   G  then neither   r 0 nor   r 1 holds  again  just do   r 0 and then flip 0 and 1 for   r 1    If   r 0 holds  then find     0   s 0       0 such that r   s 0 and the incompatibility statement holds  r   G   s 0   G  so val     0   G     val     0   G    If     1   s 1       1 such that TODO  Put everything together  since D is dense by Claim 1  find r   D   G  Therefore  by Claim 2    r 0    r 1 do not hold   Thus  r       0     1     Summary   If G is Fn         2 M   2   generic  then M   G     ZFC  there is an injection from   2 M into P         This implies M   G     ZFC   2   0     2 if we have   1 M     M   G      2 M     2 M   G    This is the goal for today s lecture   Note that our proof above is not good enough  for the statement     from Lecture 8   For all T   ZFC finite  there exists T     ZFC finite such that if M is a countable transitive model of T    then there is N   M countable transitive model of T       CH   For this  we need to look more carefully at the proof of the GMT       M   ZFC   M   G     ZFC       The proof proceeds AXIOM BY AXIOM and thus for each     ZFC  we find finite S   such that M   S   implies M   G         Let S   ZFC finite such that S proves that all relevant notions  name  value     are well defined and absolute  Then for T   ZFC finite  define      T       S         T S         Then  the proof shows       Let s prove   CH in M   G     Definition   Preserves cardinals   We say   preserves cardinals  if   G   generic  over M     is a cardinal  is absolute between M and M   G     Definition 3 23   Countable chain condition    We say   has the countable chain condition   c c c   if every antichain  note the anti   in   is countable   Theorem 3 24  Assuming that     has countable chain condition  Then    preserves cardinals   Theorem 3 25  Fn         2 M   2   has the countable chain condition   Corollary 3 26  Assuming    G is Fn         2 M   2   generic  over M  then      M   G     ZFC   2   0     2       Lemma 3 27  Assuming that   M     has countable chain condition  X   Y   M  G is   generic  over M  f   X   Y  f   M   G    Then  there is F   M such that   x   X   F   x     Y    x   X   f   x     F   x   and M     x   X   F   x   is countable   Proof of Theorem  3 24     Suppose M     is a cardinal  M   G       is not a cardinal  so there is       and f   M   G    f          f is a surjection  Apply the lemma to get F   Define R             F         Since   x   f   x     F   x    R       But M     R       0              But then M thinks that   is not a cardinal  contradiction     Now we prove the lemma   Proof  Let  F   x         y   Y     p       p       x       y       Fix   a name for f   By Definability Theorem   F   M   Then   x   X   F   x     Y follows from definition     x   X   f   x     F   x   follows from Forcing Theorem  Let s look at M   F   x    is countable  If y   F   x    let p y be such that  p y       x       y     If y   y    then p y   p y    Thus       p y   y   F   x         is an antichain   By countable chain condition  it s countalbe  So F   x   was countable     TODO  Proof of Theorem  3 25     Actually  we prove Fn   X   Y   has countable chain condition whenever Y is countable     systems form Example Sheet 3 Q33 are also called quasi disjoint families    A family of finite sets D is called a   system if there is a finite set R  called the root of the   system  such that for all D   D     D  if D   D   then D   D     R      system lemma  Any countable family of finite sets contains an uncountable   system   Take any A     uncountable and prove that it s not an antichain   If p   A  then dom   p     X finite  Consider      S       dom   p     p   A         That s an uncountable family of finite sets  so by   system lemma  find   system D   S uncountable   Let r   X finite be the root of D   Since Y is countable  there are only countably many functions q   r   Y  Since D is uncountable  by pigeonhole principle  there are p   q  such that dom   p     dom   q     D and p   r   q   r  But since dom   p     dom   q     r  p and q are compatible   So A is not an antichain     We got M   G     2   0     2   Next time  What is the size of 2   0 in M   G     Remember  LST Example Sheet 4  ZFC   2   0         What if   2 was one of the forbidden values   Remark   Obtaining M   G     2   0     2 cannot be quite as general as this proof  if M   2   0     2  then this will remain true in M   G     Remark   If G is Fn           M   2   generic  over M  then      M   G     2   0             Previous lecture  Suppose M a countable transitive model of ZFC   TODO  Question       What are the possible values for 2   0   Mentioned last lecture  not all values are possible  In particular  2   0         Definition   Cofinal   C     is cofinal     unbounded  if                 C          We can then define       cf           C     C is cofinal         Example 3 28  cf   1     1   cf         0   Lemma 3 29   K nig s Lemma      cf         Then LST Example Sheet 4 Q10 is the special case           cf       0 of K nig s Lemma  Consequence for  2 cf       2 cf     cf     2 cf    thus  2 cf         Preview of answer to      Every value not prohibited by K nig s Lemma is possible   Now    2   Definition  If   is any forcing  let      OL       A n   n           be any   sequence of antichains in     Let           OL             p     p   A n         We call these nice names   If           and   has countable chain condition  then there are at most     0 many antichains and thus at most       0     0       0     0       0 many   sequences of antichains and thus nice names   Theorem 3 30  Assuming that   M   G     x      Then  there is a nice name   such that val       G     x   Note   The Theorem does not need any assumptions about     Proof  Start with M such that val       G     x  possibly not nice    Fix n      Fix either a well ordering of    possibly using AC to get one  and build a maximal antichain A n such that   p   A n    p           TODO   Claim   val       G     val     OL   G        If n   val     O L   G    then there is p   G such that         p       OL  and then p   A n  so   p          So n   val       G        If n   val       G     then by Forcing Theorem  get q   G such that   q           Subclaim  A n   G      Indeed  by our lemma on compatibility   Example Sheet 3    get q     G such that q     p for all p   A n  Find r   q   q    Then   r          But that is in contradiction to A n being maximal   So find p   G   A n  By definition          p       OL and p   G  So  n   val     OL   G       Corollary 3 31  Assuming that     has countable chain condition  M                       0  Then  M   G     2   0       Proof  Follows directly from    a  Theorem   b  Calculation of the number of nice names      Main Application  If     Fn         2 M   2    then           2 M   Calculate in M    2   0   Hausdorff s Formula             1             1                 So in particular        1   0     1     0   0   2   0   2   0     2     1   0     2   2     So   2   0   max     2   2   0     By this calculation  if M   2   0     2  then M   G     2   0     2   Corollary 3 32  If M   CH  then M   G     2   0     2   Remark   This proof also shows that if      M   2   0     n     and G is   generic  over M where     Fn         2 M   2    then      M   G     2   0     n       Corollary  If M   CH  then M   G     2   0     n   Remark   What happens at                Fn           M   2         By general theory  M   G     2   0        but K nig s Lemma gives 2   0         1   What about the lower bound   Our theorem and counting of nice names yields      M   G     2   0         0               If M   GCH  then           0                   1       So by K nig s Lemma        0         1   Therefore  if M   GCH  then M   G     2   0         1   TODO  First limit cardinal that is a possible value of 2   0 is     1   Clearly      Fn           1 M   2   adds injection from     1 M into P        So M   G     2   0       1   Count nice names          0       1   0   If for all       1        0       1      then     1   0       1          1   0       1     f   f           1         X           1   f   f                  since   1 has no cofinal   sequence    Thus   X       1       1       1  by assumption        Thus M   G     2   0       1   TODO  What about 2   1   More on nice names   Definition   l a m b d a nice names   Generalise  nice names   to   nice  names   Let      OL     A                 be a family of   many maximal   antichains        OL               p     p   A         for         These are names for subsets of     Observe that our theorem  every A     in M   G   has a   nice name   still goes through   If   is an M cardinal such that every antichain of   has size        On Example Sheet 3  this is called the     chain condition    Let              in M   Then                             is an upper bound on the number of   nice names   Thus      M   G     2                 M       Question   Forcing with       Fn         2   2   and calculate 2   1  Assume M   GCH  Then                    2       1       0      since   has countable chain condition    So      M   G     2       2   1     0       2   1   M       Calculate     2   1   M        2   1   Hausdorff s formula   2   2   1   GCH   2     2     2       Together       M   G     2   1     2   2   0       Question   Is it possible to get PSA  i e                         2     2       without CH   In particular  can we get     2   0     2 2   1     3     First idea   Force with Fn     1     3   2      1    Yields   3 many subsets of   1    2    Still has countable chain condition  so all cardinals are preserved    3    How many   1 nice names are there         3   1     0     3   1   Hausdorff   3   2   1   GCH   3       Get  2   1     3 in M   G     Unfortunately       M   G     2   0     3       Interpret the generic object G as f       1   2 for       3   Define g       f         Claim   For          g     g      This is since      D               p     u       g     u     g       u         is still dense   So  forcing with Fn     1     3   2   gives the same situation as forcing with Fn         3   2   for 2   0  2   1   Idea      Fn   X   Y             p   dom   p     X   range   p     Y     p               Thus Fn   X   Y     Fn   X   Y     0     Consider            Fn     1     3   2     1         Properties    1    We still have M   G     2   1     3 M    2    Not clear that this forcing is preserving cardinals   First goal  What about preserving cardinals   Clearly  Fn     1     3   2     1   does not have the countable chain condition anymore   With example  38   on Example Sheet 3    we need to figure out the chain condition of Fn     1     3   2     1     We need a   system lemma for this  If   is regular   cf        and  OL is a family of sets of size     of  size      Then there is a   system  D   OL of size       This   system lemma gives with the same proof as before   Fn     1     3   2     1   has the   2 M chain condition   So  Fn     1     3   2     1   preserves cardinals     2 M   Closure  Definition   l a m b d a closed   A forcing   is called   closed  if any family   p             for       that is a descending chain               p     p       there is q such that q   p   for all         Example  Fn     1     3   2     1   is   1 cloesd    If   p     is a descending chain  then         p   is a condition in Fn     1     3   2     1      Theorem 3 33  Assuming that     is   closed  and        Then      P         M   P         M   G         Corollary 3 34  Forcing with Fn     1     3   2     1     a  Does not change P          b  Therefore preserves   1  see Example Sheet 1 and the relation between codes for countable well orders and preserving   1    Summary   Forcing with Fn     1     3   2     1   over a model of GCH gives M   G   with    1    The same cardinals  cardinals     2 preserved by   2 chain condition    1 preserved by   1 closure     2    2   1     3  standard     3    2   0     1  by corollary 1 to the closure theorem     4    Calculate number of nice names         3   1     1     3   1     3   2   1     3     2     3       Hence 2   1     3   Forcing   Property   Preservation   Arithmetic   Fn         2   2    countable chain  condition  all cardinals   2   0     2   2   1     2   Fn         3   2    countable chain  condition  all cardinals   2   0     3   2   1     3   2   2     3   Fn     1     3   2     1      1 closed    If  M   CH   then    2 chain  condition     1  preserved    Closure lemma   not yet proved           2  preserved   If  M   CH   then  2   0     1   2   1     3    Note GCH   PSA  but our model of   CH fails PSA   Question  Can we have   1   2   0   2   1   Closure lemma   Theorem  If   is   closed and        then P         M   P         M   G       closed  every descending sequence of length     has a lower bound   Proof  Let f   M   G    f       2 and assume towards contradiction that f   M    f   B      B       f   M   f   M   2         Let   be a name for f   By Forcing Theorem  there is p   G such that  p             2         B     Construct a   sequence of conditions p   TODO  TODO  The sequence   p             is defined in M  by Definability Theorem   so we can define      g         1     p     1               1         Then g   M   But now  p                 1   or  p                 0   for all     uso   p k          Hence  p         B     But  p     p       B    Contradiction     Note that while Fn   X   Y     1   is always    1 closed  the chain condition depended on the value of   1   0   The partial order Fn     1     3   2     1   has in general the   2   0     chain condition   CH implies   2   0         2  so all  cardinals are preserved   However  if 2   0     1  then there is a gap and we do not know whether TODO   If M   2   0     2  does     Fn     1 M     3 M   2     1 M       preserve   2 M   Answer  Fn             2         always adds  a surjection from     to 2     M       Application  If       2 and M   2   0     2  then Fn     1       2     1 M adds a surjection from   1 M onto P     0     M  i e    2 M  So     2 M       1 M   G      Proof of          A generic for    is  a map f             2   Define h         2   by      h               1     f             1       Claim  h is a surjection onto 2     M   If g   M  g       2  consider      D g       p                         p             1   g         1         This is dense  and thus g   range   h       Back to our question  Can we get   1   2   0   2   1   Start with M   GCH  Consider            Fn     1     3   2     1     M     and let G be   generic  over M  Consider            Fn         2   2     M   G         and let H be   generic  over M   G     Claim  M   G     H       1   2   0   2   1    M   G     H   is a forcing iteration      1      is cardinal preserving over M since M   GCH  So   n M     n M   G      2      has countable chain condition  so is cardinal preserving          n M   G     H       n M   G       n M        3    M   G     2   0     1   2   1     3  lecture 15  since M   CH     4    Then M   G     H     2   0     2   2   1     3  using a nice name analysis  we could calculate 2   1     3    This proves the claim   Important  The order of forcings matters   Suppose M   GCH              Fn         2   2     M       H     generic  over M              Fn     1     3   2     1     M   H         G     generic  over M   H     Consider M   H     G    Then      M   H     2   0     2   2   1       But that means that forcing with     will  collapse   2 M     2 M   H     Since     is   1 closed      P         M   H     P         M   H     G         In particular  M   H     G     CH  So  this order does not achieve what we want   Final Remark on Forcing CH  Assume M   2   0     2   Question  Can you obtain M   G     CH   The natural forcing would be            Fn         1 M         This collapses   1 M  it does not have the countable chain condition  but since it has size   1 M  it has the   2 M chain condition  so all cardinals     2 M are preserved   Clearly therefore       M   G       P         M       2 M     1 M   G         But  is   P         M       P         M   G       Nice names  gives upper bound of          1 M     1 M     0     2   1 M   M       That s not surprising  since any A     1 in M becomes a new subset of   in M   G   via the new bijection between   and   1 M