I have a doubt. I know C ( [0,1] , R ) ie, continuous functions from [0,1] to the real numbers is not a integral domain. But C ( R , R ), ie, continuous functions the real numbers to the real numbers is a integral domain?
Regards...
for example, let
f(x) = 0, x < 1/3
f(x) = √[(1/36) - (x-(1/2))^2], 1/3 ≤ x ≤ 2/3
f(x) = 0, x > 2/3
g(x) = 0, x < 0
g(x) = √[(1/36) - (x-(1/6))^2], 0 ≤ x ≤ 1/3
g(x) = 0, x > 1/3
then f and g are two nonzero continuous functions on R, but fg = 0, for all x.
Mmmm... so how can I prove that if C ( R , R ) is the ring continuous functions the real numbers to the real numbers. And if f is in C ( R , R ) such that f^2=f then f is the constant function f(x)=0 or is the constant function f(x)=1 ?
and if f^2 = f on the entire interval [0,1], we have the same result: f must be constant on [0,1] and equal to 0 or 1 on the entire interval.
yes, if the ring of functions on EITHER domain was an integral domain, then we could conclude that either f or f-1 was the constant 0-function.
unfortunately, that isn't the case. however, the values of f and f-1, are in an integral domain, R.
now, that means we have a function f, THAT FOR EACH VALUE OF x IN THE DOMAIN OF THE FUNCTION, is either 0, or 1.
we can't have f take on both values in any subinterval of the domain, or else (by the intermediate value theorem -since f is CONTINUOUS),
f would take on EVERY value between 0 and 1, and we know that it ONLY has the values 0 and/or 1.
so f must be constant.
this has very little to do with integral domains, and everything to do with continuity.