does such a function exist?

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January 27th 2008, 01:44 PM
hanns

does such a function exist?

I have a function h with the property
$ \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:

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

$ 2. h^{-1}\left(A\right) \ has\ not\ a\ supremum $
January 28th 2008, 03:24 AM
curvature

For the first question, let h(x)=-1/x (-1<x<0) .
January 28th 2008, 03:46 AM
curvature

I think, for the second question, the answer is NO.
January 28th 2008, 06:55 AM
hanns

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

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

is this right?
January 29th 2008, 05:03 AM
curvature

Quote:

Originally Posted by hanns

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

Then the answer is No.
If A has a supremum,(say a), but h(A) doesn't, then what is h(a+1)?

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