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 $\alpha < \omega_1$. 
$\omega_1$ is the first uncountable ordinal and it is assumed that for $\alpha, \beta < \omega_1$, $\alpha + \beta < \omega_1$. 
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 $\omega + \omega_1 < \omega_1$. 
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 $\omega_1 < \omega_1$)? 
 A: HINT: Use the definition of ordinal addition, and the fact that adding two countable ordinals is countable. 
A: You should really think of $\alpha+\beta$ as the result of "concatenating" an order of type $\alpha$ with one of type $\beta$. Formally, this is the ordinal corresponding to the well-order on $(\alpha\times\{0\})\cup(\beta\times\{1\})$, ordered by $(a,i)<(b,j)$ iff $i=j$ and $a<b$, or else $i<j$. Think of this as putting $\alpha$ dots in a row, and then $\beta$ additional dots.
To show $\omega+\omega_1=\omega_1$, it is enough to see that, in fact, already $\omega+\omega^2=\omega^2$ (and therefore $\omega+\beta=\beta$ for any $\beta\ge\omega^2$). For this, note that $1+\omega=\omega$, which should be clear from the "dots description": You have $\omega$ dots in a row, if you add an extra dot at the beginning, that is still just $\omega$ dots in a row. Now, $\omega^2$ is simply the result of replacing each of these dots with $\omega$ many. 
A: "Pictorially", the result is clear. If you draw a line, mark the points of $\omega$ on this line from left to right and then mark the points of $\omega_1$, it is more or less obvious that the resulting (well) ordered set is isomorphic to $\omega_1$.
One may explain this as follows. 
The countable ordinals start like this: $0, 1, 3,\dots ,\omega, \omega+1, \omega+2, \omega+3,\dots ,\omega+\omega=\omega\cdot 2, \omega\cdot 2+1, \omega\cdot 2+2,\omega\cdot 2+3,\dots ,\omega\cdot 2+\omega=\omega\cdot 3,\dots ,\omega\cdot n,\dots , \omega\cdot\omega, \dots $ 
If you add $\omega=\{ 0,1,2, 3,\dots \}$ on the left; this gives you the "sequence" $0,1,2,\dots , 0,1,2,\dots ,\omega, \omega+1,\omega+2,\dots ,\omega\cdot 2, \dots $. 
Now relabel your sequence in the obvious way: the second "$0$" is relabelled "$\omega$", the second "$1$" is relabelled "$\omega+1$", ..., "$\omega$" is relabelled "$\omega\cdot 2$", ..., "$\omega\cdot 2$" is relabelled "$\omega\cdot 3$" and so on. 
The transfinite sequence thus obtained is exactly $\omega_1$, i.e. $\omega+\omega_1=\omega_1$.
A more "formal" proof would consist in doing what Asaf suggests. 
A: 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 $\omega + \omega_1 = \omega_1$.
