The Internet claims that ω is the first ordinal transfinity. But what about ω-1? Isn't that a ordinal transfinity, and isn't it before ω? Kind of like ω/2?
I guess I lack an understanding of what ω is maybe? Is it equal to some set of numbers, some cardinal transfinity? Is it smaller then any infinite set of numbers? Or is it larger? Or does it even make mathematical sense to say ω<,=,or>ℝ (for example). And if it doesn't, is there some flaw in our theories of transfinity, to the extent of that some comparable numbers cannot be compared to other comparable numbers?
The ordinals have well-defined notions of successor, addition or multiplication, but not of predecessor, subtraction or division. So while it makes sense to talk about $\alpha + 1$ or $\alpha \cdot 2$ or so on, when $\alpha$ is an ordinal, it doesn't (always) make sense to talk about $\alpha - 1$ or $\frac{\alpha}{2}$, and so on.
The ordinal $\omega$ is simply the first ordinal that is greater than all the finite ordinals. There is no greatest finite ordinal, and $\omega$ is not the successor of any ordinal, so $\omega - 1$ is not defined. Likewise, there is no ordinal $\alpha$ such that $\omega = \alpha \cdot 2$, and so $\frac{\omega}{2}$ is not defined.
When formalised in set theory, the ordinals are usually encoded as von Neumann ordinals, meaning that $\omega$ is identified with the set of natural numbers.
It doesn't make sense to say $\omega < \mathbb{R}$ or $\omega = \mathbb{R}$ or $\omega > \mathbb{R}$, or whatever else, at least with the conventional meanings of the symbols $<$, $=$ and $>$, since $\omega$ is an ordinal but $\mathbb{R}$ is not. However $\mathbb{R}$ is an ordered set... you could define the notation '$\alpha \le X$', for ordinals $\alpha$ and ordered sets $X$, to mean that there is an order-embedding of $\alpha$ into $S$. In that case it would be the case that $\omega < \mathbb{R}$, since the natural numbers are embedded in the reals. But this really is just notation and not any kind of deep and subtle statement about the universe.
I don't know what you mean when you say transfinity, so I don't think I can help with your remaining questions.