Is there such an ordinal as ω-1?

Question

Sorry for the daft question, but is there such an ordinal 'ω-1' such that ω = ('ω-1')⁺? Still trying to learn set theory and ordinal arithmetic is absolutely boogling me

Answer 1

For small ordinals, it can be helpful to think about well-ordered subsets of familiar ordered sets like the real numbers. $\omega$ is order-isomorphic to $\Omega = \{ 1 - \frac{1}{n} \mid n = 1, 2 \ldots \}$. Adding $1$ to (or taking the successor of) an ordinal represented as a subset of the reals is adding a new element to the right of all the existing elements. Subtracting $1$ from (or taking the predecessor of) an ordinal represented as a subset of the reals is finding an element such that there is exactly one element to its right in the subset. Subtraction isn't always possible and $\Omega$ shows why: there are infinitely many elements of $\Omega$ to the right of any given $x \in \Omega$.