prime number theorem and prime counting function $\pi(x)$ is the prime counting function (no. of prime within x)
For the interval $(x, x + \delta x]$, $\delta > 0$, what is the smallest integer $x_{0}$ such that for any $x >= x_{0}$, $\pi(x + \delta x) - \pi(x) > 0$ is always true?
For example, Bertrand's Postulate tells us that when $\delta = 1$, the smallest integer to make the above statement true is $x_{0} = 2$.
The following result might help: one paper by Rosser and Schoenfeld gives out two inequalities about $\pi(x)$:
$\frac{x}{\log{x}}(1 + \frac{1}{2\log{x}}) < \pi(x)$, for $x>= 59$,
and $\pi(x) < \frac{x}{\log{x}}(1+ \frac{3}{2\log{x}})$, for $x>1$

I don't think that the answer above is correct. In fact, such an $ x_{0} $ does exist. It is a consequence of the Prime Number Theorem. One may refer to the Wikipedia article on Bertrand's Postulate: http://en.wikipedia.org/wiki/Bertrand%27s_Postulate. As for the original question, I'm not sure if bounds have been established for the size of $ x_{0} $, except in certain cases. Examples are also given in the Wikipedia article mentioned above. One of the references listed there, an article by Lowell Schoenfeld, gives the following result: For any $ n \geq 2010760 $, there exists a prime between $ n $ and $ \left( 1 + \frac{1}{16597} \right) n $. I believe that $ 2010760 $ is the smallest $ n_{0} \in \mathbb{N} $ corresponding to $ \delta = \frac{1}{16597} $. However, I haven't read the article yet, so please go ahead and read it to help me verify what I've written here.
