Real Analysis related to Least Upper Bound

braindead101
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Give an example of a function f for which \exists s \epsilon R P(s) ^ Q(s) ^ U(s)
P(s) is \forall x \epsilon R f(x) >= s
Q(s) is \forall t \epsilon R ( P(t) => s >= t )
U(s) is \exists y\epsilon R s.t. \forall x\epsilon R (f(x) = s =&gt; x = y)<br />
So this was actually a two part question, and this is the second part, the first part involved the function f(x) = sin(x) for which P(s)^Q(s)^U(s) could never be true.
I am not sure how to approach this, should I just think of random functions? or is there a logical way to do this.

The only thing I know is what I found from the first part of this question which is
P(s) defines the lower bound when true
Q(s) defines the greatest lower bound when true
U(s) I'm not so sure, I think something along the lines of y exist in reals, so the function must be all values of reals.. but this makes no sense as the function would not be bounded then. so I guess I am wrong.

Any help would be greatly appreciated.
 
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dear braindead101
let us consider a function f such that f(0) =0 and f(x) \neq 0 \forall x \in \mathbf{R}\backslash\{0\}. Does U(0) hold?
 
okay so, f(x) does not equal 0 for all x in reals, and then what does the last part mean? the backslash {0}
 
\mathbf{R} \backslash \{0\} is the set R with the element 0 removed. well i will give you an example of what i had in mind.
consider the function f(x)=x.
U(x) will hold for all x because f is a bijective map, so the inverse image of each y \in R will consist of a set with one element.
 
so f(x) = x but with no 0
i don't understand how p(x) and q(x) will be satisfied, arn't they not bounded in f(x) = x?
 
braindead101 said:
so f(x) = x but with no 0
i don't understand how p(x) and q(x) will be satisfied, arn't they not bounded in f(x) = x?

dear braindead101
i gave you an example of an function where U(s) holds for all s because you seemed unsure about what the meaning of U(s) is. Of course you are looking after a function which has a unique minimum (so that p(x) and q(x) will be satisfied).
 
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