Given that $\omega$ is the $ord(\mathbb{N})$, I need to prove/disprove whether there is an ordinal $\xi$ satisfying the equation $\xi$ = $\omega$+$\xi$. Here, the order given is an anti-lexicographical order, i.e., $\omega$ $\not\approxeq$ $\omega$ $+$ $1$.
I believe that there is no such ordinal, as on the left hand side, there is no maximum element, but there would be one on the right hand side, if $\xi$ was finite. I am having trouble showcasing this even when $\xi$ is an infinite ordinal.
Any help would be appreciated- thanks!
Best Answer
Consider the ordinal $\omega\cdot\omega$, then we have that $\omega+\omega\cdot\omega=\omega\cdot(1+\omega)=\omega\cdot\omega$.
Now, if $\alpha\geq\omega\cdot\omega$ is any ordinal, then there is some $\beta$ such that $\omega\cdot\omega+\beta=\alpha$ (note that $\beta$ may be equal to $\alpha$, that's fine). But now, $\omega+\alpha=\omega+\omega\cdot\omega+\beta=\omega\cdot\omega+\beta=\alpha$.
So indeed, this holds for most ordinals.