Why does $1+\omega=\omega$ with ordinal addition

Question

In my notes, it says that $1+\omega=\omega \neq \omega+1$.

I understand why the two expressions are not equal, but why does this equality holds:

$$1+\omega = \cup\{1+n:n\in\omega\}=\omega$$

Answer 1

The definition of ordinal addition says that when $\gamma$ is a limit ordinal, $\alpha+\gamma = \sup_{\beta<\gamma} (\alpha+\beta)$.

Since $\omega$ is a limit ordinal, $1 + \omega = \sup_{n<\omega} (1 + n)$.

The set $\{1 + n\mid n<\omega\}$ is the set of all finite ordinals (except $0$), so its supremum is $\omega$.

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In case it helps: You can also characterize ordinal addition as follows. If $\alpha$ and $\beta$ are ordinals, then in particular they are linearly ordered sets, and you can look at the new linear order with domain $\alpha \sqcup \beta$ ($\sqcup$ denotes disjoint union) in which every element of $\alpha$ is less than every element of $\beta$. The new linear order is still a well-order, and its order type is exactly the ordinal $\alpha+\beta$.

So you can picture $1 + \omega$ as taking $\omega$ and adding a new least element (putting $1$ to the left of $\omega$). You should easily be able to convince yourself that this has the same order type as $\omega$ (draw a picture!). On the other hand, you can picture $\omega + 1$ as taking $\omega$ and adding a new greatest element (putting $1$ to the right of $\omega$). This is different from $\omega$, since $\omega$ has no greatest element.