On 9/30/2013 11:53 PM, Arturo Magidin wrote: > On Monday, September 30, 2013 7:20:26 PM UTC-5, Hetware wrote: > >> A function is a mapping from elements of a range to elements of a >> >> domain. Often the image is required to be single valued, but >> Thomas >> >> does not stipulate that requirement. > > This "definition" does not agree with your prior assertion that a > function is a set of ordered pairs; nor does your aside about the > possibility of images being more than "single valued".
My description of a function as a set of ordered pairs was not intended to be a precise definition, though any precise definition I would use would be consistent with that description.
A function is a rule for evaluating an argument. The argument is a value in the domain, the result is a value in the range. That is, for each value in the domain (assuming functions are single-valued) there is one value in the range. The pair of values constitutes and ordered pair. That set of ordered pairs is also said to be a mapping.
> In any case, I'll wager that Thomas *does* specify that functions > must yield a unique output given any particular input (or, if he > defines functions as you did before, as sets of ordered pairs > satisfying special conditions, then one of the conditions required of > functions defined this way is that if (a,b) and (a,b') are in f, then > b=b'; i.e., that they be single valued).
Surprisingly Thomas distinguishes single-valued and multiple-valued functions. For example, y^2 = x is said to be a double-valued function of x. I would not call a one to many mapping a function.
>> If I can deterministically >> >> interpret a formal expression as such a mapping, then my >> interpretation >> >> of that formal expression satisfies the definition of a function. > > But your *interpretation* of the expression is not necessarily the > *intended* meaning of the expression. > > As I said before, if functions are set of ordered pairs (or if > functions are rules assigning to every valid input a corresponding > output), then an expression is not a function. The expression in > question is supposed to determine a function *in a particular, agreed > upon manner*.
So we are back to "that's the way is 'cause I said so". I can live with that, but the implication is that the demonstration follows from something more fundamental.
> While you are free to set up your own particular and personal > conventions, you are *not* free to impose those on the book you are > reading, which no doubt established its own intended meaning before > and that you are ignoring. >
I guess that's part of the problem. It takes a lot of work to build up rigorous definitions of even basic mathematical symbols and operations. Very few texts at the level of Thomas's Calculus go into that level of detail.
I now understand what I was missing. If I assume f(t) = (t^2-9)/(t-3) is continuous at t=3, then it is defined there. In just about everything I do, functions are assumed to be continuous unless there is some obvious coordinate, or manifold singularity. It took me a while to get the assumption of continuity out of my head.