# does such a function exist?

• Jan 27th 2008, 01:44 PM
hanns
does such a function exist?
I have a function h with the property
$\displaystyle \forall x,y \in R,\ x\leq y \rightarrow h\left(x\right) \leq h\left(y\right)$
Now A is a nonempty set which has a supremum
Are there functions such that:

$\displaystyle 1. h\left(A\right) \ has \ not\ a\ supremum$

$\displaystyle 2. h^{-1}\left(A\right) \ has\ not\ a\ supremum$
• Jan 28th 2008, 03:24 AM
curvature
For the first question, let h(x)=-1/x (-1<x<0) .
• Jan 28th 2008, 03:46 AM
curvature
I think, for the second question, the answer is NO.
• Jan 28th 2008, 06:55 AM
hanns
but -1/x is not monotonic on the whole real line
for instance $\displaystyle -1<2 \ and \ -1/-1 = 1>-1/2$

I think that the second is true for for instance $\displaystyle h^{-1}=-x$

is this right?
• Jan 29th 2008, 05:03 AM
curvature
Quote:

Originally Posted by hanns
but -1/x is not monotonic on the whole real line
for instance $\displaystyle -1<2 \ and \ -1/-1 = 1>-1/2$