Prove $\omega + \omega_1 = \omega_1$

Question

I am assuming that $\omega_1$ is the first uncountable ordinal and I'm using ordinal arithmetic.

I have so far that if $\alpha$ and $\beta$ are ordinals, then $\alpha + \beta$ = sup{$\alpha + \gamma$ | $\gamma < \beta\}$ where $\beta$ is a limit ordinal.

So since $\omega$ and $\omega_1$ are both ordinals, can I say that $\omega + \omega_1$ = sup{$\omega + \gamma$ | $\gamma < \omega_1 \}$?

And then I'm not sure where to go with this.

Thanks in advanced.

Answer 1

Here are a couple of nice facts that should help you.

• First of all, the sum of two countable ordinals is countable (for the simple reason that the union of two countable sets is countable). This should give a nice upper bound for $\omega + \omega_1$.

• Secondly, note that $\alpha + \beta \geq \max \{ \alpha , \beta \}$ for all $\alpha, \beta$, which gives a nice lower bound for $\omega + \omega_1$.