A perhaps silly question about > 0

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September 9th 2009, 12:01 AM
Luca

A perhaps silly question about > 0

Now and then I see domains defined as in this example:
1) $x \geq \epsilon > 0$
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) $x > 0$
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
September 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 $x\geq\epsilon>0$ .
September 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

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

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

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

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

Exactly as you suggested, in such case $|z| < 1$ 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 $|z|$ closer and closer to 1. Whereas chosing a fixed value for $\epsilon > 0$ would allow us to find an index $N_{\delta}$ , function only of a $\delta>0$ , at which we could stop adding the terms $z^n$ so that

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

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

Thanks for the tip.

Luca