f:R-->R and f '(x)<= a < 0 for any x
show that f is onto R
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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.
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Does anyone know what is wrong with this proof if anything? Thank you
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.
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
$\arctan(-x)$ 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.
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?
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?
I'm pretty sure that monotonicity and continuity imply the existence of an inverse of f.
Given that proving f is onto is trivial. $\forall y \in Y\;\; f\left(f^{-1}(y)\right)=y$
Isn't this easier than your current approach?