# A perhaps silly question about > 0

• Sep 9th 2009, 12:01 AM
Luca
A perhaps silly question about > 0
Now and then I see domains defined as in this example:
1) $\displaystyle x \geq \epsilon > 0$
for any epsilon greater than zero, no matter how small.

I would be tempted to believe that the above cannot be much different from
2) $\displaystyle x > 0$
as in my understanding that already excludes the case x=0.
Yet, there must be some subtle reason why some author prefers to explicitly write it as in 1)

Can it be just a matter of conventions ? or am I missing some fundamental definition ?
Thanks
Luca
• Sep 9th 2009, 04:53 AM
Bruno J.
Good question. Obviously, this will depend on the actual issue, but I can give an example of why that might happen.

Very often, a function will converge uniformly everywhere except in a neighbourhood of some point, say 0, where perhaps the function is undefined and around which it does not behave well. Then we cannot say it converges uniformly for x>0, but we can say that it converges uniformly for $\displaystyle x\geq\epsilon>0$.
• Sep 9th 2009, 08:09 AM
Luca
The geometric series might be a good example
Thanks Bruno,
I believe I have finally got it.
And perhaps, the geometric series might be a good example.

Let us consider the geometric series

$\displaystyle \sum_{n=1}^\infty z^n$

z being a complex number. It is known that said series is convergent within $\displaystyle |z| < 1$.
However, for claiming in addition that the convergence is uniform it is probably more correct to define a disk of uniform convergence as

$\displaystyle |z| \leq 1-\epsilon < 1$

for any $\displaystyle \epsilon > 0$, no matter how small.

Exactly as you suggested, in such case $\displaystyle |z| < 1$ would not be sufficient for uniform convergence, as it would result in the need to add more and more terms to get sufficient accuracy as we choose $\displaystyle |z|$ closer and closer to 1. Whereas chosing a fixed value for $\displaystyle \epsilon > 0$ would allow us to find an index $\displaystyle N_{\delta}$, function only of a $\displaystyle \delta>0$, at which we could stop adding the terms $\displaystyle z^n$ so that

$\displaystyle \left|\sum_{n=1}^{N_{\delta}} z^n -\frac{1}{1-z}\right|<\delta$

is valid for any $\displaystyle z$ in the domain $\displaystyle |z| \leq 1-\epsilon < 1$

Thanks for the tip.

Luca