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Topic: Is (t^2-9)/(t-3) defined at t=3?
Replies: 166   Last Post: Oct 30, 2013 9:41 AM

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Hetware

Posts: 148
Registered: 4/13/13
Re: Is (t^2-9)/(t-3) defined at t=3?
Posted: Oct 20, 2013 10:57 AM
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On 10/20/2013 6:40 AM, Peter Percival wrote:
> Hetware wrote:
>

>> What is your opinion of the definition given as follows:
>>
>> Let f(x) be continuous over R, and f(x) = x/x for all x in R where x/x
>> is a determinate form?
>>
>> I believe f(x) to be consistently and uniquely defined for all x in R,
>> under that specification.

>
> So the domain of f is R\{0} and f(x) = 1. Since f isn't defined at 0 it
> cannot be continuous there.


Asserting that f(x) is continuous means that f(0) is defined, and f(0) =
limit[s/s, s->0]. See below.

> It makes no sense to say let f be continuous, and _then_ define it.
> Define f first and then prove it to be continuous if it is so.


"Let f(x) be continuous over R" is part of the definition.

> If continuous over R means continuous at every real, then it is false of f.
>


I'm not sure the question as to whether a function can be defined as
continuous is directly relevant to my suggestion that the continuity of
f(x)=x/x is ambiguous, but I do find the question interesting.

NB: I will admit to equivocating on the suggestion that proposing a
previously defined function is continuous has the same effect as
defining it as continuous.

Can I write: let f(x) for all x in R be a function defined by f(x) = 1
if x is real, and f(x) = 0 if x is rational?

Of course I can! I just did.

The point is that I can make a statement that has the structure of a
definition of a function, but which does not define a function because
it fails to satisfy the requirements of a function. So any proposed
function must be validated. That would apply to asserting continuity as
part of the definition. So your insistence that we must _prove_
continuity rather than assert it is addressed because the continuity of
the function must be verified in order to validate the definition.

With that in mind, here is what I mean by "Let f(x) be a continuous
function over R":

Given some function r(x) which over R has no more than a finite number
of removable discontinuities and no irremovable discontinuities, and
f(x) = r(x) where r(x) is determinate then f(x) = limit[r(s),s->x].

NB: That is not very well stated, but it's a start.

I submit that this is a valid definition of a continuous function: "For
all x in R, if x/x is determinate let f(x) = x/x , else let f(x) =
limit[s/s,s->x]." . Do you agree?

Now, using my meaning of "Let f(x) be a continuous function over R" we
state the following:

Let f(x) be a continuous function over R such that f(x) = x/x, where x/x
is determinate.

That is the same as saying: let f(x) = x/x if x/x is determinate, else
f(x) = limit[s/s,s->x].

And that could just as well be stated: let f(x) = limit[s/s,s->x].

I vaguely recall encountering that means of "patching" a function r(x)
having a finite number of discontinuities. The author instructed the
reader to treat such a function as identical to r(x) = limit[r(s),s->x]
for all practical purposes. If I can recall where I read that, I will
provide a citation. And no, it was not intended as a rigorous definition.

In sum, defining a function as continuous is just an abbreviated means
of "patching" a discontinuous function in the definition.

It seems to me there should be a term which distinguished a function
with at most a finite number of discontinuities from a function with one
or more irremovable discontinuities. I propose "indeterminate
continuity" for the former.

I'm not even going to address the case of a function with an infinite
number of removable discontinuities. It makes my head hurt.


Date Subject Author
9/28/13
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Hetware
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