reserve
  X for non empty set,
  FX for Filter of X,
  SFX for Subset-Family of X;

theorem
  for X be infinite set holds
  the set of all X\A where A is finite Subset of X is Filter of X
  proof
    let X be infinite set;
    set FF=the set of all X\A where A is finite Subset of X;
    now
      let x be object;
      assume x in FF;
      then consider a1 be finite Subset of X such that
A1:   x=X\a1;
      thus x in bool X by A1;
    end;
    then FF c= bool X;
    then reconsider FF as non empty Subset-Family of X by Th28;
A2: not {} in FF
    proof
      assume {} in FF;
      then consider a be finite Subset of X such that
A3:   {}=X\a;
      X c= a by A3,XBOOLE_1:37;
      hence contradiction;
    end;
A4: for Y1,Y2 be Subset of X st Y1 in FF & Y2 in FF holds
    Y1/\Y2 in FF
    proof
      let Y1,Y2 be Subset of X such that
A5:   Y1 in FF and
A6:   Y2 in FF;
      consider a1 be finite Subset of X such that
A7:   Y1=X\a1 by A5;
      consider a2 be finite Subset of X such that
A8:   Y2=X\a2 by A6;
      Y1/\Y2=X\(a1\/a2) by A7,A8,XBOOLE_1:53;
      hence thesis;
    end;
    for Y1,Y2 be Subset of X st Y1 in FF & Y1 c= Y2 holds Y2 in FF
    proof
      let Y1,Y2 be Subset of X such that
A9:   Y1 in FF and
A10:  Y1 c= Y2;
      consider a1 be finite Subset of X such that
A11:  Y1=X\a1 by A9;
      X\Y2 c= X\(X\a1) by A10,A11,XBOOLE_1:34;
      then
A12:  X\Y2 c= X/\a1 by XBOOLE_1:48;
      X\(X\Y2)=X/\Y2 & X/\Y2 c= Y2 by XBOOLE_1:17,XBOOLE_1:48;
      then
      X\(X\Y2) = Y2 by XBOOLE_1:28;
      hence thesis by A12;
    end;
    hence thesis by A2,A4,CARD_FIL:def 1;
  end;
