Random interarrival times (poisson process) In their monograph "Queues", Cox and Smith state (paraphrased - this is p5):
In interval $(t, \Delta t)$ the probability of no arrivals in a completely random process is $1 - \alpha \Delta t + o(\Delta t) $, for one arrival $ \alpha \Delta t + o(\Delta t)$ and for more than one arrival $o(\Delta t)$ where $ \alpha $ is the mean rate of arrival.
I cannot follow this... here is my thinking - we take $N$ to be the probability of no arrivals, $W$ to be the probability of one arrival, $Z$ to be the probability of more than one arrival, and $A$ to be the probability of any arrivals.
So $1 = N + W + Z = N + A $
By my understanding of Cox and Smith:
$1 = 1 - \alpha \Delta t + o(\Delta t) + \alpha \Delta t + o(\Delta t)+ o(\Delta t)$ $= 1 + 3o(\Delta t) $ which is surely nonsense.
So, what have I got wrong here?
 A: Yes, Profesor Rubin's answer points the right direction. You might want to look up  little-o notation to get an introduction to what is going on here. (NB, it is worth being precise about which little-o is meant because there are several related uses, especially in probability, where you also run into more general ideas such as stochastic order symbols, where we don't just control the "growth-rate" of a function but (in some sense) of a series of distribution. We don't need those here, however!) For the current purposes we can non-rigourously take $o(x)$ to be "any  function" such that $$\lim_{x\to 0} \frac{o(x)}{x}=0$$
Now, let $X(t)$ count the number of arrivals in the time $(0,t]$ (and let us set $X(0):=0$ for simplicity). Then we may rewrite your statement as
$$
P[X(t+\Delta t)-X(t)] = 1-\alpha \Delta t +o(\Delta t)
$$
which implies 
$$
\begin{align*}
\lim_{\Delta t\to 0}\frac{P[X(t+\Delta t)-X(t)=0]}{\Delta t}&=\lim_{\Delta t\to 0}\frac{(1-\alpha \Delta t +o(\Delta t))-1 }{\Delta t}\\
&=\lim_{\Delta t\to 0}\frac{-\alpha \Delta t +o(\Delta t)}{\Delta t}\\
&=\lim_{\Delta t\to 0}\frac{-\alpha \Delta t}{\Delta t} + \lim_{\Delta t\to 0} \frac{o(\Delta t)}{\Delta t}\\
&=-\alpha + 0\\
&=-\alpha \\
\end{align*}
$$
Similar calculations show that
$$
\lim_{\Delta t\to 0}\frac{P[X(t+\Delta t)-X(t)=1]}{\Delta t}=\alpha
$$
and
$$
\lim_{\Delta t\to 0}\frac{P[X(t+\Delta t)-X(t)>1]}{\Delta t}=0
$$
A possibly surprising fact is the description you have given here is nearly enough to show that $X(t)\sim\operatorname{Poisson}(\alpha t)$ - all you need in addition is that the increments of this process are independent. I suspect that your queueing theory book would have the derivation.
Back to your point though, it bears thinking carefully about the litte-o-notation! in particular, when I said any function above, I was serious -we don't assume the particular form of any given use of $o(x)$. So, for example, in general $o(x)\ne o(x)$, and $o(x)-o(x)=o(x)$, $o(x)=o(x)$, and $2o(x)=o(x)$, but $xo(x)=o(x^2)$ - and, which we used above, $0.o(x)=0$.
You can prove all these and more from the definition of little-o notation given in the wikipedia link above, and it's worthwhile if you want to make sure you use it right!
BTW, I wouldn't really call this a queuing theory question as such - the little o-notation is way more general!
A: From Professor Paul Rubin:
"Your error lies in assuming that all o(dt) thingies are the same, which they are not (or at least not necessarily). The notation o(dt) just means some term that dies (converges to zero) faster than dt can get to zero. The term can be positive, negative or both (meaning it switches sign as dt changes). So, as an example, 1 = (1 – dt + dt^2) + (dt – dt^3) + (dt^3 – dt^2) = [1 - dt + o(dt)] + [dt + o(dt)] + [o(dt)] whre the first o(dt) is dt^2, the second is -dt^3, the third is dt^3 – dt^2, and they all converge to 0 faster than dt does.
"O() and o() notation is in a way the mathematical version of an ellipsis (“I’m leaving this out because it is too scary/boring/lengthy to contemplate”), with an additional restriction that the thing being ignored is unimportant because it vanishes when you start taking limits."
