On 10/19/2013 11:30 PM, Arturo Magidin wrote: > On Saturday, October 19, 2013 7:42:48 PM UTC-5, Hetware wrote:
>> >>> Hey, Mr. Pot. Have you met Ms Kettle?
Are those your parents?
You are wrong to think that I am somehow emotionally attached to being right in my original assertions. As I have already stated, I do not conform my understanding to the assertions of others unless I understand and can validate those assertions.
If I fully understood the subtleties involved, I would not be spending nearly as much time discussing this matter. I am seeking understanding, not vindication.
>> What convention is that? > > As I explained eariler: > > The convention that when the domain of a function is not explicitly > stated, then the domain is taken to be the "natural domain" of the > function. > > The "natural domain" of a function that is given via a formula *and > no other specification* is taken to be the set of all real numbers > for which the formula, as given, makes sense and yields a real > number.
Given some real-valued function of real-valued arguments, defined everywhere on R, is it true that the function is either continuous, or not continuous?
Let P(f) be the proposition that f(x) is continuous for all x in R. Then !P(f) means f(x) is discontinuous for some x in R.
Does P(f) have meaning? Is it an admissible mathematical statement?
> You keep talking about assumptions of continuity, assumptions of > this, assumptions of that, authors trying to "bait" you, etc. But you > keep ignoring the conventions. > >>> Rather than admit this, you are going out of your way to try to >>> prove >> >>> that you were right all along. >> >> >> >> The proposition P(f) that "f(x) is continuous" appears to have >> meaning. >> >> So either P or !P. > > And so we get back to your self-justifying dance. Sorry, I don't > tango with the willfully ignorant. >
Let P(f) <=> f(x) is continuous for all x in R. Let D(f) <=> f(x) has a finite value for all x in R. Let L(f) <=> limit[f(s), s->x] exists for all x in R. Let Q(f) <=> f(x) = limit[f(s), s->x]
P <=> D & L & Q Q => D
Let f(x) be in R for all x in R other than x=0, f(0) = 0/0 and limit[f(s), s->x] exist for all x in R.
D(f) => F L(f) => T Q(f) => F !P(f) <= F & T & F
Let g(x) be a continuous function on R such that g(x) = f(x) when f(x) is a real number.
D(g) => T L(g) => T Q(g) => T P(g) <= T & T & T
So Quasi was correct. I can define a function g(x) to be continuous, but it is not the same function as the function f(x) used to define g(x) everywhere f(x) is defined.
