
theorem Th16:
  for FT1,FT2 being non empty RelStr, A being Subset of FT1,B
being Subset of FT2, f being Function of FT1, FT2 st f is_continuous 0 & B=f.:A
  holds f.:(A^b) c= B^b
proof
  let FT1,FT2 be non empty RelStr,A be Subset of FT1, B be Subset of FT2, f be
  Function of FT1, FT2;
  assume that
A1: f is_continuous 0 and
A2: B=f.:A;
  thus f.:(A^b) c= B^b
  proof
    let y be object;
    assume y in f.:(A^b);
    then consider x being object such that
A3: x in dom f and
A4: x in A^b and
A5: y=f.x by FUNCT_1:def 6;
    reconsider x0=x as Element of FT1 by A3;
    reconsider y0=y as Element of FT2 by A3,A5,FUNCT_2:5;
    f.:( U_FT(x0,0)) c= U_FT(y0,0) by A1,A5;
    then f.:(U_FT x0) c= U_FT(y0,0) by FINTOPO3:47;
    then f.:(U_FT x0) c= U_FT y0 by FINTOPO3:47;
    then
A6: f.:((U_FT x0) /\ A) c= (f.:((U_FT x0)))/\ (f.:A) & (f.:((U_FT x0)))/\
    (f.:A) c= (U_FT y0) /\ (f.:A) by RELAT_1:121,XBOOLE_1:26;
    ex z being Element of FT1 st z=x & U_FT z meets A by A4;
    then
A7: U_FT x0 /\ A<>{} by XBOOLE_0:def 7;
    dom f=the carrier of FT1 by FUNCT_2:def 1;
    then f.:((U_FT x0) /\ A)<>{} by A7,RELAT_1:119;
    then (U_FT y0) /\ (f.:A)<>{} by A6;
    then U_FT y0 meets B by A2,XBOOLE_0:def 7;
    hence thesis;
  end;
end;
