Re: Reasons to vote Motion 27: NO
Jürgen and P1788
I overlooked your 2 Aug email, and replied yesterday without taking it into account. Apologies. However your response doesn't change my view, for the reasons below.
On 2 Aug 2011, at 08:23, J. Wolff von Gudenberg wrote:
>> And, guys, you have an FTDIA. Your FTIA is Theorem 2, and your FTDIA is Theorem 3. But your Theorem 3 is either FALSE, or means something other than its obvious meaning, see my 19 July email (below) to you, Jürgen, which according to my records you haven't answered.
> Theorem 3 means that the expression given as an extention of a function is calculated by a DET following(1).
>
> Theorem 3 reformulated
> Given a DET f~ for an extension of f. The mechanism (1) then calculates a decorated interval (e,d) where d<= di for all di appearing in the DET
> That gives a proof that only functions with properties described by the decoration d have been applied
Already this is false, because the decorations are specified in Definition 1 to be logically disjoint. E.g., the predicate of "def" includes the clause "f is not continuous on xx". I had taken this to be one of the "minor technical slips" in Motion 27, but you argue inconsistently so it is necessary to ask if this is what you really intend.
> if d = saf f is everywhere defined and continuous
Yes
> if d = def f is everywhere defined
Yes, but it doesn't mean "the decoration of f" is def. (d is the computed decoration, and "the decoration of f" is the true decoration.)
f might be continuous as well, so
> if d = def the decoration of f is either def or saf
> (which means f is everywhere defined, as you say)
What if d = con? Then the decoration of f could be con or def or saf according to your definition.
But what you don't seem to have taken into account is that the decoration of f could also be ndf!
Consider f(x)=sqrt(-1-x*x) on xx=[-2,2].
We have sqrt(-1 - [-2,2]*[-2,2]) = sqrt(-1=[-4,4]) = sqrt([-5,3]), which gives d = con, but
the true decoration is ndf!!!
So your formula does NOT give a lower bound for the true decoration of f. I would suggest that, if we continue discussing this, you should provide mathematically sound text. There are simply gaps that
should not be there in the first place.
This is the actual state of affairs ASSUMING your decorations are exactly as specified in Definition 1:
Computed d | Possible values of true decoration of f
saf | saf
def | def or saf
con | ndf or con or def or saf
ndf | ndf (Nothing else is possible, which is
| another thing not made clear by Theorem 3)
So your "obvious" Theorem 3 comes nowhere near describing the truth!
John