• Feb 26th 2014, 08:56 AM
computerproof
f:R-->R and f '(x)<= a < 0 for any x

show that f is onto R
---------------------------------------------------------------------------------------------------------------------
Proof: since f '(x) < 0, we have f(x) a monotone decreasing function

since f '(x) exists, we know f(x) must be continuous for any x in R

the cardinality of two like sets is the same so: |R| = |R|

Since the cardinality of the domain and range is the same, proving onto amounts to proving that f:R-->R is one-to-one

But f(a)>f(b) for any a,b in R

Thus, f:a-->f(a) and f:b-->f(b), and

f(a) is not equal to f(b) whenever a is not equal to b

which is to say f(x) is one-to-one for any x in R

Since the cardinality of the domain equals the cardinality of the range (i.e. |R|=|R|) and f is one-to-one,

f(x) must be ONTO.
--------------------------------------------------------------------------------------------------

Does anyone know what is wrong with this proof if anything? Thank you
• Feb 26th 2014, 09:31 AM
Plato
Quote:

Originally Posted by computerproof
f:R-->R and f '(x)<= a < 0 for any x

show that f is onto R

Consider $f(x)=-e^x-x$. Is that a counter-example?
• Feb 26th 2014, 09:41 AM
johng
Hi,
The error in your "proof" is where you assert that if two sets A and B have the same cardinality, a function from A to B is onto if and only if f is one to one. Wrong, as Plato's example shows. The statement about onto and one to one is true only for finite sets A and B.
• Feb 26th 2014, 09:51 AM
computerproof
I don't see how it is a counterexample being that f '(x) is always negative forcing f(x) to be monotone decreasing. since f is defined for every possible value in the domain R, and f(x) is one-to-one: whenever a does not equal b we have f(a) does not equal f(b). With teh cardinality of the domain and range being the same I think this proves onto....but something still doesn't seem right...lol

johng....i wrote this response before i had seen what you had written.....

so the cardinality rule i state applies only to finite sets...ok

can you explain the prposed counterexample....can't seem to wrap my mind around it
• Feb 26th 2014, 10:05 AM
romsek
$\arctan(-x)$ is another

it maps $(-\infty, \infty)$ to $\left(\frac{\pi}{2}, -\frac{\pi}{2}\right)$ yet is monotone decreasing over it's entire domain.

So it meets your criterion of continuous and negative derivative but is not onto.
• Feb 26th 2014, 10:10 AM
computerproof
but arctan(-x) is bounded....since f '(x)<a<0, theis means f(x) < ax.....but since x can be any real number, f(x) can take on any value in R which is to say our f(x) is not bounded.....so i need to show f(x) is unbounded?
• Feb 26th 2014, 10:12 AM
computerproof
can I use the fundamental theorem of calculs to say that this: f '(x) <=a<0

implies this: f(x) <= ax

(i basically integrated both sides
• Feb 26th 2014, 10:15 AM
romsek
you're right. I didn't catch the upper bound on f'
• Feb 26th 2014, 10:22 AM
computerproof
I feel unsure of myself when i use theorems because they do not always perfectly line up. for example the fundamental theorem of calculus is stated only for a finite interval....can i use it in cases like my question above where we are dealing with R, an infinite interval?
• Feb 26th 2014, 03:01 PM
romsek
Given that proving f is onto is trivial. $\forall y \in Y\;\; f\left(f^{-1}(y)\right)=y$