reserve N for Cardinal;
reserve M for Aleph;
reserve X for non empty set;
reserve Y,Z,Z1,Z2,Y1,Y2,Y3,Y4 for Subset of X;
reserve S for Subset-Family of X;
reserve x for set;
reserve F,Uf for Filter of X;
reserve S for non empty Subset-Family of X;
reserve I for Ideal of X;
reserve S,S1 for Subset-Family of X;
reserve FS for non empty Subset of Filters(X);
reserve X for infinite set;
reserve Y,Y1,Y2,Z for Subset of X;
reserve F,Uf for Filter of X;
reserve x for Element of X;

theorem Th33:
  M is measurable implies M is regular
proof
A1: cf M c= M by CARD_5:def 1;
  assume M is measurable;
  then consider Uf being Filter of M such that
A2: Uf is_complete_with M and
A3: Uf is non principal being_ultrafilter;
  assume not M is regular;
  then cf M <> M by CARD_5:def 3;
  then
A4: cf M in M by A1,CARD_1:3;
  then consider xi being Ordinal-Sequence such that
A5: dom xi = cf M and
A6: rng xi c= M and
  xi is increasing and
A7: M = sup xi and
  xi is Cardinal-Function and
  not 0 in rng xi by CARD_5:29;
  M = sup rng xi by A7,ORDINAL2:def 5;
  then
A8: M = union rng xi by A6,Th32;
  Uf is_complete_with card M by A2;
  then
A9: Uf is uniform by A3,Th23;
A10: rng xi c= dual Uf
  proof
    let X be object such that
A11: X in rng xi;
    reconsider X1=X as Subset of M by A6,A11,ORDINAL1:def 2;
A12: card X1 in M by A6,A11,CARD_1:9;
    not X1 in Uf
    proof
      assume X1 in Uf;
      then card X1 = card M by A9;
      then card X1 = M;
      hence contradiction by A12;
    end;
    hence thesis by A3,Th22;
  end;
  card rng xi c= card dom xi by CARD_2:61;
  then card rng xi c= cf M by A5;
  then
A13: card rng xi in M by A4,ORDINAL1:12;
  dual Uf is_complete_with M by A2,Th12;
  then union rng xi in dual Uf by A8,A13,A10,ZFMISC_1:2;
  hence contradiction by A8,Def2;
end;
