How to interpret the little-$o$ notation in this definition of the Poisson process? In the book Probability and Random Processes by Grimmett and Stirzaker, a Poisson process is defined to be any process $(N(t))_{t\in[0,\infty)}$ with values in $\mathbb{N}_0$ satisfying following three properties:
$\hspace{20pt}$(a) $N(0) = 0$ and $s<t\Rightarrow N(s)\leq N(t)$
$\hspace{20pt}$(b) $\mathbb{P}\left(N(t+h)=n+m|\hbox{ }N(t) =n\right) =\begin{cases}
o(h) & \text{if } m>1,\\
\lambda h + o(h) &\text{if } m=1,\\
1 - \lambda h + o(h) & \text{if }m=0\\
\end{cases}$
$\hspace{20pt}$(c) if $s<t$ then $N(t)-N(s)$ is independent of the times of emissions on $[0,s]$.
I am having problems with understanding the meaning $o(h)$ in (b), so my question is:

How do we interpret $o(h)$ in this definition?

I know $o(h)$ in the classical sense is supposed to be just a function such that $\lim_{h\to 0}\frac{o(h)}h=0$. But here, $o$ is used three times and I suspect it has a different meaning each time. (Since otherwise, we would necessarily have $o\equiv0$.)
So, I did some thinking and reinterpreted (b) to read:
$\hspace{20pt}$(b') There exist functions $o_1, o_2, o_3$, such that for $i=1,2,3$:$$\lim_{h\to 0}\frac{o_i(h)}h=0$$
$\hspace{38pt}$and
$$\mathbb{P}\left(N(t+h)=n+m|\hbox{ }N(t) =n\right) =\begin{cases}
o_1(h) & \text{if } m>1,\\
\lambda h + o_2(h) &\text{if } m=1,\\
1 - \lambda h + o_3(h) & \text{if }m=0.\\
\end{cases}$$
But I am completely confused about the order of quantifiers in this statement (which is why I left this rewording a bit vague). Should the functions $o_1,o_2,o_3$ be the same for all $t,n$ and $m>1$? (That is: should we take a separate function $o_m$ for each m? Or even a separate function $o_{t,n,m}$ for each triple $(t,n,m)$?)

Does this even matter or are these definitions miraculously equivalent?

I would really like to know what exactly it is this definition is trying to define. Thanks in advance for any helpful suggestions.
 A: Yes, the convention for using little-o notation is that every $o(\cdots)$ in an equation can stand for a different function that is eventually smaller than the identity.
Also, usually the precise function is allowed to depend in arbitrary ways on anything that is not mentioned in the "\cdots".
Consider, for example the not unusual definition of the derivative of $f(x)$ at $x_0$ being the (unique) $\lambda$ such that $f(x_0+h)=f(x_0)+\lambda h + o(h)$. Here it is implicit that the $o(\cdots)$ function can depend on $x_0$ but not on $h$.
A: Once you've found exact formulas for the Poisson process, it could help your intuition to  go back and write the functions explicitly. I will re-index the $o$ functions to match $m$
as follows:
$$\mathbb{P}\left(N(t+h)=n+m|\hbox{ }N(t) =n\right) =\begin{cases}
1 - \lambda h + o_0(h) & \text{if }m=0\\
\lambda h + o_1(h) &\text{if } m=1,\\
o_m(h) & \text{if } m>1,\\
\end{cases}
$$
From the explicit formulas, we have 
$$\begin{eqnarray*}
o_0(h)&=&e^{-\lambda h}-1+\lambda h\\
o_1(h)&=&\lambda h(e^{-\lambda h}-1)\\
o_m(h)&=&(\lambda h)^m\, e^{-\lambda h}/m! \ \mbox{ for } m>1\\
\end{eqnarray*}
$$
As you see, these functions depend on $m$ and $\lambda$, but not $n$ or $t$. 
