Why is it that $\omega + \omega_1 = \omega_1$? I have seen some other questions about the proofs around ordinals and $\omega$ and $\omega_1$, but some of the answers have confused me.

An ordinal $\alpha$ is countable if $\alpha &lt; \omega_1$. 

$\omega_1$ is the first uncountable ordinal and it is assumed that for $\alpha, \beta &lt; \omega_1$, $\alpha + \beta &lt; \omega_1$. 

Now, my initial thought, similar to the answers in the 'Prove $\omega + \omega_1 = \omega_1$' style of questions is that we substitute $\omega$ and $\omega_1$ as $\alpha$ and $\beta$ respectively. Then I would obtain $\omega + \omega_1 &lt; \omega_1$. 

Similarly, I could substitute again into $\alpha + \beta \geq max \{\alpha, \beta \}$ to get $\omega + \omega_1 \geq max \{\omega, \omega_1 \} = \omega_1$. So we have an upper and lower bound and can claim $\omega + \omega_1 = \omega_1$. 

But my question is, how does this work since $\omega_1$ is uncountable (and so we cannot say $\omega_1 &lt; \omega_1$)?
HINT: Use the definition of ordinal addition, and the fact that adding two countable ordinals is countable.
Every ordinal number can be expressed as a multiple of $\omega^2$ plus an ordinal number less than $\omega^2$. If you take any multiple of $\omega^2$ and left add $\omega$, you get back the exact same ordinal number. If you take any countable multiple of $\omega^2$ and right add any ordinal number less than $\omega^2$, you still get a countable ordinal. Thus, $\omega_1$ is a multiple of $\omega^2$ and so $\omega + \omega_1 = \omega_1$.