> You really see no difference between > > Let f be a continuous, twice differentiable function... > > and > > Let f be defined by <some formula written here> and > f is continuous because I say so. > > ? > >
The notion of "formula" must be taken as extremely broad. Any means of determining a set of ordered pairs (x, f(x)) will be a formula.
The following is Mathematica code for a function F[x, C, r] where x is some real value, C is a Boolean value representing the assertion of continuity, and r is some expression mapping elements of the domain to elements of the range.
(* if continuity of F not asserted return r[x]*) F[x_, C_, r_] := r[x] /; ! C
(* else if r[x] is continuous return r[x].*) F[x_, C_, r_] := r[x] /; ContinuousQ[r, x]
(* else if r[x] is a removable discontinuity return Limit[r[t],t->x].*) F[x_, C_, r_] := Limit[r[t], t -> x] /; RemovableQ[r, x]
(* else asserting F[x,True,r] leads to a contradiction.*) F[x_, C_, r_] := False
(*The following are examples of using F[x,C,r] to define function of the form (x, f(x)) where x and f(x) are real numbers.*) r[x_] := x/x f[x_] := F[x, True, r] Print["r[x_]:=x/x"] Print["f[x_]:=F[x,True, r]"]
So given the mapping r[x_] := x/x we define a function f[x_] := F[x, True, r] which uses the definition of continuity and r[x] to find a suitable value for f[x]. If r[x] results in an indeterminate form, f[x] has a rule which returns the value of Limit[r[t], t -> x]. If r[x] is not a suitable candidate, that is to say it has one or more irremovable discontinuities, then f[x] returns False, meaning that the definition resulted in a contradiction.
Notice that not asserting continuity gives a different result. In fact, g[x_]:=F[x,False,r] is just g[x]=r[x].
Is the proposition 'the mapping m is continuous' meaningful?