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Of course, this is an application of the theorem: AnyGiven :

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

Clearly $H$ is bounded above by $b$. Now you want to show that $H\ne\emptyset$.

Apply the

$\exists c\in (a,b)$ 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 $H\ne\emptyset~?$

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So according to your proof :Of course, this is an application of the theorem: Anynon-emptyset 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 theintermediate value theoremtwice.

$\exists c\in (a,b)$ 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 $H\ne\emptyset~?$

$\exists t\in (a,c)$ 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?

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?

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

$\sup(H)$ must exist.

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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 thecompleteness property

$\sup(H)$ must exist.

I was reading a proof for the intermediate value theorem and the author starts the proof by writing:

Let H be a set such that : for all xε(a,b) , f(x)<d.

Then he goes on saying : Clearly H has a supremum,since H is bounded above by b and $H\neq\emptyset$

Surely in proving the intermediate value theorem we cannot use the intermediate value for part of its proof.

That is why i asked for any other suggestions meaning other proof.

No ,no .maybe you misunderstood me.

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

As written, the OP looks like an exercise that goes with that section a textbook.

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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 theintermediate value theorem?

As written, the OP looks like an exercise that goes with that section a textbook.

Next time give all the information!

I thought it was an easy problem just escaping my attention .

Sorry for that.

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