# Nested Intervals and IVT

• Nov 2nd 2010, 06:19 PM
zebra2147
Nested Intervals and IVT
I could really use some help. Not really sure how to get started.

Suppose $f(a,b)< 0$ and $f(c,d)>0$. Construct nested intervals $[a_{n},b_{n}]$ and $[c_{n},d_{n}]$ such that $f(a_{n},c_{n})\leq 0$ and $f(b_{n},d_{n})>0$. Then show $f(x_{0},y_{0})=0$ if $\bigcap _{n=1}^\infty [a_{n},b_{n}]=\{x_{0}\}$ and $\bigcap _{n=1}^\infty [c_{n},d_{n}]=\{y_{0}\}$.
• Nov 2nd 2010, 06:25 PM
Drexel28
Quote:

Originally Posted by zebra2147
I could really use some help. Not really sure how to get started.

Suppose $f(a,b)< 0$ and $f(c,d)>0$. Construct nested intervals $[a_{n},b_{n}]$ and $[c_{n},d_{n}]$ such that $f(a_{n},c_{n})\leq 0$ and $f(b_{n},d_{n})>0$. Then show $f(x_{0},y_{0})=0$ if $\bigcap _{n=1}^\infty [a_{n},b_{n}]=\{x_{0}\}$ and $\bigcap _{n=1}^\infty [c_{n},d_{n}]=\{y_{0}\}$.

What is this $f$ just some continuous map $f:[0,1]\times[0,1]\to\mathbb{R}$? What are $a,b,c,d$? I think you need to type up the full question.
• Nov 3rd 2010, 04:50 AM
zebra2147
$f$ is a continuous function on some interval $I$ where $a,b,c,d\in I$.
• Nov 3rd 2010, 07:51 AM
Drexel28
Quote:

Originally Posted by zebra2147
$f$ is a continuous function on some interval $I$ where $a,b,c,d\in I$.

Ok? Then what does $f(x,y)$ mean? Unless you meant $f\left((a,b)\right)$ but then the inequality signs used are meaningless.
• Nov 14th 2010, 10:05 AM
zebra2147
There was a typo to this problem initially. It should read...
Suppose $f(a,c)<0$ and $f(b,d)>0$.

I'm guessing we need to use the Intermediate Value Theorem to show that if there exists $f(x_0,y_0)=0$ then $f(a,c). And then somehow use this to show that $\bigcap _{n=1}^\infty [a_{n},b_{n}]=\{x_{0}\}$ and $\bigcap _{n=1}^\infty [c_{n},d_{n}]=\{y_{0}\}$??
• Nov 14th 2010, 10:14 AM
Drexel28
Quote:

Originally Posted by zebra2147
There was a typo to this problem initially. It should read...
Suppose $f(a,c)<0$ and $f(b,d)>0$.

I'm guessing we need to use the Intermediate Value Theorem to show that if there exists $f(x_0,y_0)=0$ then $f(a,c). And then somehow use this to show that $\bigcap _{n=1}^\infty [a_{n},b_{n}]=\{x_{0}\}$ and $\bigcap _{n=1}^\infty [c_{n},d_{n}]=\{y_{0}\}$??

You have yet to answer my question. What does $f(a,c)<0$ mean?
• Nov 14th 2010, 02:55 PM
zebra2147
Well, my professor never told us but judging by the position of the proof in my notes I think we are trying to help prove the following theorem:

If $I$ and $J$ are intervals and $f:I\times J\rightarrow \mathbb{R}$ is continuous, then the range of $f$ is an interval.

So is it possible that $f(a,b)$ is an interval??
• Nov 14th 2010, 03:12 PM
Drexel28
Quote:

Originally Posted by zebra2147
Well, my professor never told us but judging by the position of the proof in my notes I think we are trying to help prove the following theorem:

If $I$ and $J$ are intervals and $f:I\times J\rightarrow \mathbb{R}$ is continuous, then the range of $f$ is an interval.

So is it possible that $f(a,b)$ is an interval??

But isn't it more likely given what you just said that you're suppose to consider $(a,b)\in I\times J$?
• Nov 14th 2010, 05:22 PM
zebra2147
Ok well if we consider $(a,b)\in I\times J$ could you help me get started?
• Nov 14th 2010, 05:23 PM
Drexel28
Quote:

Originally Posted by zebra2147
Ok well if we consider $(a,b)\in I\times J$ could you help me get started?

I'm sorry, no. Until I understand what's going on, I can't say anything intelligible.