Is every infinite countable limit ordinal of form $\beta+ \omega$?

Question

Prove or disprove: For every infinite limit ordinal $\alpha < \omega_1$, there is an ordinal $\beta$ such that $\alpha = \beta + \omega$.

I know that there are no different infinite ordinals that are simultaneously additively and multiplicatively commmutative. Does that helps?

Answer 1

Take α=ω·ω and note that for every x ∈ α, there exists y ∈ α such that x < y and the interval [x,y] is infinite.

It is clear that this fails for β + ω (Take x = β, then for every y ∈ β+ω, [x,y] is finite).

Hence α $\neq$ β + ω for any ordinal β.

Answer 2

If $\omega^2 = \beta + \omega$, then $\beta< \omega^2 = \{a\omega+b: a,b\in\omega\}$.

But then $\beta = a\omega+b$ for some $a, b\in \omega$.

Therefore $\beta+\omega = (a+1)\omega<\omega^2$. Contradiction.