# First fundamental theorem, question about continuity

• August 19th 2010, 09:01 PM
Mollier
First fundamental theorem, question about continuity
Hi,

I was reading the proof for the first fundamental theorem in Apostol, and one part of it says the following.

Continuity of $f$ at $x$ tells us that, if $\epsilon$ is given, there is a positive $\delta$ such that

$|f(t)-f(x)|<\frac{1}{2}\epsilon$

whenever

$x-\delta < t < x+\delta$

I do not see how $|f(t)-f(x)|$ must be less than $\frac{1}{2}\epsilon$. I do see why it has to be less than $\epsilon$ though..

Any hints?

Thanks.
• August 19th 2010, 09:06 PM
Defunkt
It's simply a matter of formalism - the author chose to take $\epsilon_0 = \frac{1}{2} \epsilon$ and used $\epsilon_0$ in the definition of continuity.
• August 19th 2010, 09:08 PM
Mollier
So he just chooses a smaller neighborhood?
• August 19th 2010, 09:13 PM
Defunkt
Well, it doesn't really matter. The proof would've worked as well, had he taken $\epsilon$ instead of $\frac{1}{2} \epsilon$. The only difference would be that in the end, he would have $| \text{something} | < 2 \epsilon$, and by taking $\epsilon_0 = \frac{1}{2} \epsilon$, he will end up with $| \text{something} |< \epsilon$.
• August 19th 2010, 09:16 PM
Mollier
Ah, he's really thinking ahead then. Must be a good chess player :)
Thanks mate.
• August 20th 2010, 03:53 AM
HallsofIvy
Quote:

Originally Posted by Mollier
Ah, he's really thinking ahead then. Must be a good chess player :)
Thanks mate.

Actually, most proofs like this are developed by taking " $\epsilon$", seeing at the end that it would have been better to use " $\frac{1}{2}\epsilon$", then going back and changing it- which you can't do in chess.