
MHB Master
#1
February 15th, 2020,
04:25
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:
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

February 15th, 2020 04:25
# ADS
Circuit advertisement

#2
February 15th, 2020,
10:07
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 $x$.

MHB Master
#3
February 16th, 2020,
22:50
Thread Author
Originally Posted by
GJA
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:
Were you indeed invoking something like what Apostol calls the signpreserving property of continuous functions?
Thanks again for your help ...
Peter

#4
February 17th, 2020,
00:54
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.