# 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 $\omega_1 < \omega_1$)?

HINT: Use the definition of ordinal addition, and the fact that adding two countable ordinals is countable.

• $\alpha + \beta = sup \{\alpha + \gamma | \gamma < \beta \}$. $\omega + \omega_1 = sup \{\omega + \gamma | \gamma < \omega_1 \}$. And since $\gamma < \omega_1$ it is countable, and the supremum is the same as $\bigcup \{\omega + \gamma | \gamma < \omega_1 \}$, which is a union of countable ordinals. Is that right? – Math Student 91 Mar 16 '14 at 17:25
• Yes. And what can be the supremum of uncountably many different countable ordinals? – Asaf Karagila Mar 16 '14 at 17:27
• would this be $\omega_1$? – Math Student 91 Mar 16 '14 at 17:49
• What could it be? – Asaf Karagila Mar 16 '14 at 18:45
• Don't guess. Look at the definitions. What could be the supremum of a set of countable ordinals? – Asaf Karagila Mar 17 '14 at 11:30

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.

"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.