Re: Cantor Confusion

In article <ssXIUBVEGUrFFwxp@xxxxxxxxxxxxxxxxxxxxxxxxxx> Andy Smith <Andy@xxxxxxxxxxxxxxxxxxxx> writes:
....
That function cannot be made continuous at 0 at all. It is an essential
discontinuity.

Doubtless you are right, it is very badly behaved, but I would like to
know the basis for the assertion.

Simply because lim{x -> 0} sin(pi/x) does not exist. And a function is
continuous only if the limit exists and if the limit is also the function
value on the point. A fucntion is essentially discontinuous at some point
if there is *no* value that makes it continuous.

[ about exp(-1/x^2) ]
Yes, I agree, you can't define the derivatives, only the function value
at the point, my mistake. But I don't see the distinction between this
situation and sin(pi/x) - in one case, the function can apparently be
defined arbitrarily at x = 0, and in the other it has to be 0?

No, in both cases you can set it arbitrarily. However, in the case of
sin(pi/x) the limit to x = 0 does not exist. In the case of exp(-1/x^2)
that limit *does* exist, and it is equal to 0.

If you
define exp(-1/x^2) to be other than 0 at x = 0, then the derivatives
will be discontinous (and infinite),

No, when you define it as anything different from 0, the derivative is
undefined at x = 0.

and such a defined function will
have no valid Taylor expansion, but so what? You are happy to introduce
a discontinuity and an arbitrary value at x=0 on sin(pi/x) ?

Go ahead. But that does not make sin(pi/x) well defined at x = 0. There
is *no* way to define that function for x = 0 so that the function is
continuous at x = 0. The case is different for g(x) = exp(-1/x^2).
Even when you define g(0) = 0, *all* the derivatives at x = 0 are 0.
So there is no valid Taylor expansion around x = 0. Or perhaps it is
valid, but all the coefficients are 0. To quote something you wrote
earlier:
but you CAN define a function f by
e.g. Taylor expansion in x about some x0, such that f(x) =
exp(1/x^2) for all x!=0,

There is no way to do that. If you do a Taylor expansion around some
x0, that Taylor expansion will have a radius of convergence, i.e. a
circle around x0 where it converges (assuming complex analysis). That
circle never includes singular points of the function, and in this
case the circle would be the circle around x0 that has the origin on
its circumference. So the Taylor series would not work for -x0.
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