I'm reading a proof in graph theory, and the method is to consider two cases of graphs separately: those with maximum degree $\Delta < 2^{o(\sqrt{\log n})}$, and those with $\Delta \geq 2^{\Omega(\sqrt{\log n})}$.
So, it seems as though (whatever the author means by this notation) that these are two mutually exclusive cases. More generally, I suppose, $f(x)<o(g(x))$ and $f(x) \geq \Omega (g(x))$ are mutually exclusive circumstances, in the author's notation.
So, may I ask, what precisely is meant by this notation? I know the definition of $f(x)=o(g(x))$, and I know that it implies $f(x)=O(g(x))$. I am only used to writing things like $f(x)=O(g(x))$, and never something like $f(x)< O(g(x))$.
EDIT: Here is the book, if anyone wants more context: https://www.cs.bgu.ac.il/~elkinm/book.pdf
I am referring to the paragraph before Lemma 10.6 on pg113, and the paragraph before Theorem 10.7 on pg114.
About the original problem. In general I don't think that these two cases cover all cases there are. Let us denote $\log_2 \Delta = a_n$. Then, they consider two cases:
$$ \frac{a_n}{\sqrt{\ln n}} \to 0 \quad \text{and} \quad a_n \geq c \sqrt{\ln n} \quad \text{for some $c > 0$.} $$
In principle it could be possible that $a_n / \sqrt{\ln n} = 1$ if $n$ is even and $\tfrac{1}{n}$ when $n$ is odd, implying that the sequence does not belong to any of the classes described above. Perhaps some additional information on the quantity being studied might imply that cases like the one I just described are impossible.