# Intermediate value theorem 2

#### solakis

##### Active member
If:

f is a real value function continuous over [a,b] ,a<b and f(a)<f(b)

,then prove that for all η,
f(a)<η<f(b) ,there exists an ξ ,such that : f(ξ)=η

Proof.

1)Let S be a set of all xε[a,b] such that :f(x)<η.

Since a belongs to S and b is an upper bound then S has a Supremum

2)Let ξ= SupS

3) Will show that : a) $f(\xi)\geq\eta$ ,b) $f(\xi)\leq\eta$ and hence : $f(\xi)=\eta$

3a) By the definition of supremum, for every positive ε, there is a number x' of S with
$\xi-\epsilon\leq x'\leq\xi$ .
For this x', f(x')<η.
Since f is contiguous at ξ, $f(\xi)\leq\eta$ which is : 3a

3b) Any x greater than ξ is not in S and so $f(x)\geq \eta$.By continuity ,f(ξ) is the limit of f(x) as x tends to ξ through values greater than ξ ,and so :$f(\xi)\geq\eta$

I have no idea how he gets $f(\xi)\geq \eta$ in part 3b and $f(\xi)\leq\eta$ in part 3a

#### Amer

##### Active member
Re: intermediate value theorem 2

If:

f is a real value function continuous over [a,b] ,a<b and f(a)<f(b)

,then prove that for all η,
f(a)<η<f(b) ,there exists an ξ ,such that : f(ξ)=η

Proof.

1)Let S be a set of all xε[a,b] such that :f(x)<η.

Since a belongs to S and b is an upper bound then S has a Supremum

2)Let ξ= SupS

3) Will show that : a) $f(\xi)\geq\eta$ ,b) $f(\xi)\leq\eta$ and hence : $f(\xi)=\eta$

3a) By the definition of supremum, for every positive ε, there is a number x' of S with
$\xi-\epsilon\leq x'\leq\xi$ .
For this x', f(x')<η.
Since f is contiguous at ξ, $f(\xi)\leq\eta$ which is : 3a

3b) Any x greater than ξ is not in S and so $f(x)\geq \eta$.By continuity ,f(ξ) is the limit of f(x) as x tends to ξ through values greater than ξ ,and so :$f(\xi)\geq\eta$

I have no idea how he gets $f(\xi)\geq \eta$ in part 3b and $f(\xi)\leq\eta$ in part 3a
3a) $$f(\xi) \leq \eta$$
Pick $$\epsilon = \eta - f(\xi)$$ as you should know
f is continuous at ξ that means for any $$\epsilon >0$$ there exist $$\delta >0$$
such that whenever $$\mid x - \xi \mid < \delta$$ we have $$\mid f(x) - f(\xi) \mid < \epsilon$$
remember that epsilon>0 so if you choose a wrong epsilon the criteria for the continuity will be affected, back to our proof after we choose epsilon sub it
$$\mid f(x) - f(\xi) \mid < \eta - f(\xi)$$

$$-\eta + f(\xi) < f(x) - f(\xi) < \eta - f(\xi)$$

$$f(x) - f(\xi) < \eta - f(\xi) \Rightarrow f(x) < \eta$$ for

$$\mid x - \xi \mid < \delta \Rightarrow \xi - \delta < x < \xi + \delta$$

but $$\xi$$ is an upper bound for S, $$\xi < \xi +\frac{\delta}{2}$$
and
$$\xi + \frac{\delta}{2} \in S$$ since
$$f\left(\xi + \frac{\delta}{2}\right) < \eta$$ but $$\xi$$ is an upper bound for S contradiction!! so $$\eta - f(\xi)$$ is not positive so still two choices less than or equal zero

Lets see if $$f(\xi) > \eta$$
f is continuous at $$\xi$$ , now pick $$\epsilon = f(\xi) - \eta$$
there exist delta

$$\mid f(x) - f(\xi) \mid < f(\xi) - \eta$$ for $$\mid x - \xi \mid < \delta$$

$$-f(\xi) + \eta < f(x) - f(\xi)$$

$$\eta < f(x)$$ for $$\xi - \delta < x < \xi + \delta$$

$$\xi - \delta$$ is in S the definition of the upper bound which means $$f \left(\xi - \delta \right)$$ should be less than $$\eta$$ not larger contradiction so our epsilon is not positive

This idea from Wiki see this

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#### solakis

##### Active member
Re: intermediate value theorem 2

3a) $$f(\xi) \leq \eta$$
Pick $$\epsilon = \eta - f(\xi)$$ as you should know
f is continuous at ξ that means for any $$\epsilon >0$$ there exist $$\delta >0$$
such that whenever $$\mid x - \xi \mid < \delta$$ we have $$\mid f(x) - f(\xi) \mid < \epsilon$$
remember that epsilon>0 so if you choose a wrong epsilon the criteria for the continuity will be affected, back to our proof after we choose epsilon sub it
$$\mid f(x) - f(\xi) \mid < \eta - f(\xi)$$

$$-\eta + f(\xi) < f(x) - f(\xi) < \eta - f(\xi)$$

$$f(x) - f(\xi) < \eta - f(\xi) \Rightarrow f(x) < \eta$$ for

$$\mid x - \xi \mid < \delta \Rightarrow \xi - \delta < x < \xi + \delta$$

but $$\xi$$ is an upper bound for S, $$\xi < \xi +\frac{\delta}{2}$$
and
$$\xi + \frac{\delta}{2} \in S$$ since
$$f\left(\xi + \frac{\delta}{2}\right) < \eta$$ but $$\xi$$ is an upper bound for S contradiction!! so $$\eta - f(\xi)$$ is not positive so still two choices less than or equal zero

Lets see if $$f(\xi) > \eta$$
f is continuous at $$\xi$$ , now pick $$\epsilon = f(\xi) - \eta$$
there exist delta

$$\mid f(x) - f(\xi) \mid < f(\xi) - \eta$$ for $$\mid x - \xi \mid < \delta$$

$$-f(\xi) + \eta < f(x) - f(\xi)$$

$$\eta < f(x)$$ for $$\xi - \delta < x < \xi + \delta$$

$$\xi - \delta$$ is in S the definition of the upper bound which means $$f \left(\xi - \delta \right)$$ should be less than $$\eta$$ not larger contradiction so our epsilon is not positive

This idea from Wiki see this
The proof you proposed is completely different from the proof i wrote in my OP.

Sorry but i am not asking for a proof of the IVT but an explanation of the OP's proof