User:80.98.179.160

I'm possibly the one who thinks that \(2\omega\ne\omega\), but if yes, here's the proof:

Assume that \(2\omega=\omega\).

This follows \(\omega=0\), but since NO transfinite ordinal is 0, thus \(2\omega\ne\omega\).

And \(1+\omega\ne\omega\), due to the Axioms being violated if \(1+\omega=\omega\).