# Real Analysis related to Least Upper Bound

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 $$\exists y\epsilon R s.t. \forall x\epsilon R (f(x) = s => x = y)$$
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.

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?

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?