What does $O\left(\frac{1}{\log\log T}\right)$ mean?

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I've started to work through this paper by Levinson (1975). The abstract says

"Let $s = \sigma + it$. For any complex a, all but $O\left(\frac{1}{\log \log T}\right)$ of the roots of $\zeta(s) = a$ in $T < t < 2T$ lie in $\left|\frac{1}{2}-\sigma\right| < \frac{(\log \log T)^2}{\log T}$. ..."

I think I understand what most of this is saying. Here's a diagram I have sketched which expresses my current understanding:

pic1

Basically if we choose a real value of $T$ then for $s=\sigma+it$, where $T<t<2T$, "almost all" (see the paper title) of the zeros of $\zeta(s)=a$ will lie within $\frac{(\log\log T)^2}{\log T}$ of $\sigma=1/2$, i.e. almost all of the zeros, $s_i=\sigma_i+t_i$, will be such that $\left|\frac{1}{2}-\sigma_i\right|<\frac{(\log\log T)^2}{\log T}$.

The bit which is confusing me is "...all but $O\left(\frac{1}{\log\log T}\right)$ of the roots...". What does this mean exactly? Suppose for a moment there were 100 such roots and we choose $T=10$, then the passage would read "...all but $O(\approx 1.19899)$ of the roots..." I don't know what this means. Does it mean $1.19899\times 100=119.899>100$ roots don't lie within the said region. I think not. And what if the number of roots is actually infinite (which I suspect it is).

I am used to using the Big-O notation like $$O(x^2)=c_1x^2+c_2x^3+\cdots,$$ but not that described in the abstract clearly.

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Big-O notation describes how fast a function grows -- this usually only includes the fastest growing part of the function (but may include more if deemed signficant): $O(n^2)$ could be any thing where $n^2$ is the fastest growing component of the function The function may include any other components such as $1$, $log(n)$ or $n$ or $\sqrt{n}$ or any combination which doesn't grow faster than $n^2$, multiplied with any constant factors. Things like $n^3$ and $n^2\log(n)$ are not allowed.

Constant factors are left out, except when comparing to other functions where the constants are the interesting parts, but that's very unusual.