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Intermediate Value Theorem ...Silva, Theorem 4.2.1 ... ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
I am reading Cesar E. Silva's book entitled "Invitation to Real Analysis" ... and am focused on Chapter 4: Continuous Functions ...

I need help to clarify an aspect of the proof of Theorem 4.2.1, the Intermediate Value Theorem ... ...

Theorem 4.2.1 and its related Corollary read as follows:



Silva - 1 - Theorem 4.2.1 & Corollary 4.2.3 ... PART 1.png
Silva - 2 - Theorem 4.2.1 & Corollary 4.2.3 ... PART 2 .png



In the above proof by Silva, we read the following:

" ... ... So there exists \(\displaystyle x\) with \(\displaystyle b \gt x \gt \beta\) and such that \(\displaystyle f(x) \lt 0\) ... ... "


My question is as follows:

How can we be sure that \(\displaystyle f(x) \lt 0\) given \(\displaystyle x\) with \(\displaystyle b \gt x \gt \beta\) ... indeed how do we show rigorously that for \(\displaystyle x\) such that \(\displaystyle b \gt x \gt \beta\) we have \(\displaystyle f(x) \lt 0\) ...


Help will be much appreciated ...

Peter
 

GJA

Well-known member
MHB Math Scholar
Jan 16, 2013
260
Hi Peter ,

Since $\beta<b$ and $\delta >0$, there are values of $x$ such that $|x-\beta|<\delta$ and $\beta < x< b$. The key here is to remember that $|x-\beta| <\delta$. By the continuity argument, $f(x)<0$ for all such $x$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
Hi Peter ,

Since $\beta<b$ and $\delta >0$, there are values of $x$ such that $|x-\beta|<\delta$ and $\beta < x< b$. The key here is to remember that $|x-\beta| <\delta$. By the continuity argument, $f(x)<0$ for all such $x$.





Thanks for the help GJA!

At first I struggled with what you meant by ... " By the continuity argument, $f(x)<0$ for all such $x$ ... "

But then I found Apostol Theorem 3.7 (Calculus Vol. 1, page 143) which reads as follows:


Ap[ostol - Calculus - Theorem 3.7 .png


Were you indeed invoking something like what Apostol calls the sign-preserving property of continuous functions?


Thanks again for your help ...

Peter
 

GJA

Well-known member
MHB Math Scholar
Jan 16, 2013
260
Hi Peter ,

Happy to help!

I wasn't quoting that purposely, though it is true. In fact, it's essentially what the author is proving by their choice of epsilon.

What I meant was: $|f(x)-f(\beta)|<\epsilon\,\Longrightarrow\, f(x)<\epsilon + f(\beta)<0.$

Hope this helps clear up the confusion on my earlier post.