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Of course, this is an application of the theorem: Any non-emptyy set of real numbers that is bounded above has a least upper bound.Given :
1) f : [a,b] => R
2) f is continuous over [a,b]
3) f(a)<d<f(b)
4) a<b
Then prove that the following,set: H ={x: xε(a,b),f(x)<d} has a supremum
So according to your proof :Of course, this is an application of the theorem: Any non-empty set of real numbers that is bounded above has a least upper bound.
Clearly $H$ is bounded above by $b$. Now you want to show that $H\ne\emptyset$.
Apply the intermediate value theorem twice.
$\exists c\in (a,b)$ such that $f(c)=d$ or $f(a)<f(c)=d<f(b)$.
This means that $f(a)<\frac{d+f(a)}{2}<d$ so $\exists t\in (a,c)$ such that $f(t)=\frac{d+f(a)}{2}$.
How does that show that $H\ne\emptyset~?$
So according to your proof :
$\exists t\in (a,c)$ such that $f(t)=\frac{d+f(a)}{2}<d \Longrightarrow t\in H\Longrightarrow H\neq\emptyset$.
But unfortunately i found the above at the beginning of a proof for the intermediate value theorem.
So any other suggestions for the above?
No ,no .maybe you misunderstood me.There is no need for other suggestions. That is the proof.
$H$ is a non-empty set that is bounded above by $b$, by the completeness property
$\sup(H)$ must exist.
No ,no .maybe you misunderstood me.
I do apologize,but i did not realized that by simply asking for another solution to the problem could cause such a mental distress.You bet I misunderstood your OP, and it is your fault.
It the of OP, why did you not make it clear that you were proving the intermediate value theorem?
As written, the OP looks like an exercise that goes with that section a textbook.
Next time give all the information!