# Thread: Closed and Bounded Intervals

1. ## Closed and Bounded Intervals

Let $f:[0,1]\rightarrow \mathbb{R}$ be a function that maps closed and bounded intervals onto closed and bounded intervals, hence if $0\leq a, then for some real numbers $c\leq d, f([a,b])=[c,d]$.

1)Prove that there is an $x_{0}$ contained in $[0,1]$such that for all $x$ contained in $[0,1], f(x)\leq f(x_{0}).$
2)Prove the range of $f$ has the intermediate value propery.
3)Give an example showing $f$ not needing to be continuous.

My attempts/observations...
1) It would appear that we are looking for an upper bound for $f$? I'm not reall too sure how to start that one though...

2) Well, possibly we can define $f(0)$ and $f(1)$ and then use that for $0\leq a, $c\leq d, f([a,b])=[c,d]$. Then show that $[c,d]$ is between $f(0) and f(1)$???

3)I think I can probably take a better guess once I have a better understand of 1) and 2)

Any help would be greatly appreciated.

2. Originally Posted by zebra2147
Let $f:[0,1]\rightarrow \mathbb{R}$ be a function that maps closed and bounded intervals onto closed and bounded intervals, hence if $0\leq a, then for some real numbers $c\leq d, f([a,b])=[c,d]$.

1)Prove that there is an $x_{0}$ contained in $[0,1]$such that for all $x$ contained in $[0,1], f(x)\leq f(x_{0}).$
2)Prove the range of $f$ has the intermediate value propery.
3)Give an example showing $f$ not needing to be continuous.

My attempts/observations...
1) It would appear that we are looking for an upper bound for $f$? I'm not reall too sure how to start that one though...

2) Well, possibly we can define $f(0)$ and $f(1)$ and then use that for $0\leq a, $c\leq d, f([a,b])=[c,d]$. Then show that $[c,d]$ is between $f(0) and f(1)$???

3)I think I can probably take a better guess once I have a better understand of 1) and 2)

Any help would be greatly appreciated.
1) So you know that $f\left(\left[0,1\right]\right)=[c,d]$, and so $\sup f\left([0,1]\right)=\sup [c,d]=d$ but $d\in f\left([a,b]\right)$...sooo

2) What does the range of $f$ having the IVP mean. Isn't he IVP a property of functions?

3. 1) well since $d=sup{f([0,1])$ and $[a,b]\in [0,1]$ then $d=sup(f[a,b])$???

2)Well, in my notes I have that the range of $f$ has the intermediate value property if $f$ is continuous on some interval. So could we somehow prove that $f$ is continuous on $[0,1]$ to show that the range of $f$ has the IV property?

4. Originally Posted by zebra2147
1) well since $d=sup{f([0,1])$ and $[a,b]\in [0,1]$ then $d=sup(f[a,b])$???

2)Well, in my notes I have that the range of $f$ has the intermediate value property if $f$ is continuous on some interval. So could we somehow prove that $f$ is continuous on $[0,1]$ to show that the range of $f$ has the IV property?
Ok, now I'm confused. Where does this $[a,b]$ come into play, and the IVP is another word fro continuity??

5. Ok... How about this...
1) Since $d=sup[0,1]$ then we have that $d$ is an upper bound. Thus, by definition, for any $x\in [0,1]$, $f(x)\leq f(d)$
2) I was just saying that a theorem in my notes states that if some function $f$ is continuous then the range of $f$ has the intermediate value property so I thought we could maybe use that.
But... in order for the range of $f$ to have the intermediate value property... here is my guess. For some $a where $a,b,t\in f(x)$, $f(x_1)=a$, $f(x_2)=b$ then there must be a $f(x_3)$ such that $f(x_3)=t$ for some $x_1. But the more I look at what I just said I'm guessing this only works for strictly increasing continuous functions so I'm sure that's not right.

6. Originally Posted by zebra2147
Ok... How about this...
1) Since $d=sup[0,1]$ then we have that $d$ is an upper bound. Thus, by definition, for any $x\in [0,1]$, $f(x)\leq f(d)$
2) I was just saying that a theorem in my notes states that if some function $f$ is continuous then the range of $f$ has the intermediate value property so I thought we could maybe use that.
But... in order for the range of $f$ to have the intermediate value property... here is my guess. For some $a where $a,b,t\in f(x)$, $f(x_1)=a$, $f(x_2)=b$ then there must be a $f(x_3)$ such that $f(x_3)=t$ for some $x_1. But the more I look at what I just said I'm guessing this only works for strictly increasing continuous functions so I'm sure that's not right.
1) looks ok to me as far as I understand the question

2) I think really what you're saying (your book has bad terminology) is that the function $f:[a,b]\to\mathbb{R}$ has the IVP. But, isn't this obvious since it's image is an interval?

7. Well, I believe you if that is what you are telling me. So I guess that is not the direction that we need to go. So could you point me in the right direction since I'm guessing my last post was way off?

8. Originally Posted by zebra2147
Well, I believe you if that is what you are telling me. So I guess that is not the direction that we need to go. So could you point me in the right direction since I'm guessing my last post was way off?
To be honest I am still missing exactly what IVP means in this case. Can you EXPLICITLY write out the definition?

9. For a hint at a discontinuous function with the given property, think about sin(1/x).

10. The only definition that I have for the IVP is the following:
Let $f$ be continuous on some interval $I$. If there are $a, such that $f(a)<0$ and $f(b)>0$, then there exists a $c\in [a,b]$, such that $f(c)=0$.

This is the only defintion that I have. My professor didn't explicitly define the IVP for the range of a function.

11. Could it mean that if $a,b\in f([0,1])$ then $[a,b]\subset f([0,1])$?

12. So are you saying the Intermediate Value Property would come into play since if $a,b\in f([0,1])$ then $[a,b]\subset f([0,1])$. Thus since $[a,b]$ is an interval then there must be some other point cointained in $[a,b]$? Thus, the range of $f$ has the intermediate value property.