Nested Intervals and IVT

**zebra2147** · November 2nd 2010, 06:19 PM

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 $\bigcap _{n=1}^\infty [c_{n},d_{n}]=\{y_{0}\}$ .

**Drexel28** · November 2nd 2010, 06:25 PM

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.

**zebra2147** · November 3rd 2010, 04:50 AM

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

**Drexel28** · November 3rd 2010, 07:51 AM

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.

**zebra2147** · November 14th 2010, 10:05 AM

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 $f(a,c)<f(x_0,y_0)<f(b,d)$ . And then somehow use this to show that $\bigcap _{n=1}^\infty [a_{n},b_{n}]=\{x_{0}\}$ and $\bigcap _{n=1}^\infty [c_{n},d_{n}]=\{y_{0}\}$ ??

**Drexel28** · November 14th 2010, 10:14 AM

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 $f(a,c)<f(x_0,y_0)<f(b,d)$ . 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?

**zebra2147** · November 14th 2010, 02:55 PM

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??

**Drexel28** · November 14th 2010, 03:12 PM

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 $(a,b)\in I\times J$ ?

**zebra2147** · November 14th 2010, 05:22 PM

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

**Drexel28** · November 14th 2010, 05:23 PM

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.

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