Re: Cantor Confusion

Andy Smith wrote:
Dave Seaman wrote:
No, sin(pi/x) is undefined at x = 0.

You can define a function f: R -> R by

f(x) = sin(pi/x), if x != 0,
= 0, if x = 0.

and in that case, it's true that f(x) = 0 (but
because the definition of
f says so, not because of antisymmetry. You can
deduce antisymmetry from
the definition, not the other way around.

If f(x) is such that at any eta, f(eta) = -f(-eta),
this would imply that f(0) = 0?

If f is defined at zero and you know that f(-x)=-f(x), then setting x to
zero gives that f(0)=-f(0), which implies f(0)=0. But, if you have a
function that is not defined at zero, then it isn't true that
f(x)=-f(-x) for all x. It may be true for the x in the domain of f, but
this by itself doesn't force you to increase the domain. In other words,
if f isn't defined at zero, then writing f(0) is meaningless.

If f(0) was anything other than 0, the antisymmetry would be destroyed?

f doesn't have to be anything at zero. Functions have whatever domain
you give them when you define them.

How could something that is antisymmetric not be other than 0
at its mirror point - even if it is multivalued, it will retain
antisymmetry in inverting x ?

Functions are not multivalued.

According to Wikipedia, an odd function has to have domain all of R. So,
according to this definition, sin(pi/x) is not odd, since it isn't
defined at zero. The function f that Dave Seaman defined above is odd.

Alll this is a bit "angels on a pin", but is ultimately important
I think from a philosophical perspective on the nature of infinity.

You seem to have a funny idea about math. Mathematical objects are as we
define them. You seem to want to argue in reverse.

--
David Marcus
.

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