Prove or disprove: Every limit ordinal has the form α+ω for some ordinal α.
I needed help completing this proof and to see if I was on the right track. My idea was to use induction on α. Suppose β=α+ω for some ordinal α and let $$A=(β_1,β_2,…)$$ the set of all limit ordinals.
Base case: Suppose $β=ω$, then for all nϵω will do since $$β=ω=n+ω$$
Induction hypothesis: Assume that $$β_i=α_i+ω$$
for all $i<n$.
This is where i get stuck, im not sure if i complete this using induction. But if anyone can offer suggestion that would help a lot. Thanks
Claim. Every limit ordinal $\alpha$ is of the form $$ \alpha = \omega \cdot \beta $$ for some unique ordinal $\beta > 0$. Furthermore $\alpha$ can be written as $$ \alpha = \gamma + \omega $$ if and only if $\beta$, as above, is a successor ordinal.
It's easy to prove this claim via induction on $\alpha$. The case that $\alpha$ is a limit of limit ordinals is trivial. Otherwise there is some largest limit ordinal $\bar{\alpha} < \alpha$ to which you can apply the induction hypothesis. (Note that $\bar{\alpha} + \omega = \alpha$.)