Comparing ordinals

Question

I'm trying to teach myself set theory, and know that for any two ordinals $α$, $β$, exactly one of $α ∈ β$, $α = β$, $β ∈ α$ hold.

The notes I'm working have an exercise which asks me to determine which of these holds when

(i) $α = (ω + 1).2$, $ β = 2.(ω + 1)$

(ii) $ α = (ω + 1).ω$, $ β = ω.(ω + 1)$

but I haven't seen any examples where this has been done, and having messed around with the definitions of ordinal arithmetic haven't been able to put the pairs in forms I can easily compare.

Any help you could offer would be really appreciated.

Answer 1

I want to offer a more "order-oriented" approach:

$(\omega + 1) \cdot 2$ is the order where we replace every point in $2$ by a disjoint copy of $\omega +1$, so that's $\omega$, a maximum so far, $\omega$, total maximum; the maximum after the first copy can be seen as part of the second copy of $\omega$, so in total we get $\omega+\omega+1$.

$2\cdot (\omega+1)$ is the order where we replace each point of $\omega+1$ by a copy of $2$, so a doubling of points. The initial $\omega$ gets replaced by a new copy of $\omega$ , the final maximum gets doubled, so that's in fact $\omega+2$ as an ordinal, and this is an initial segment of the previous order from the first paragraph.

Try now to visualize why $(\omega+1) \cdot \omega$ is smaller than $\omega \cdot (\omega+1)$. E.g. $(\omega+1) \cdot \omega$ is $\omega$ in which we replace each point by $\omega+1$. but the final max in $\omega+1$ is "eaten up" by the following copy of $\omega+1$ as its minimum, so the result is just $\omega^2$ in the end.

Answer 2

Some useful facts to keep in mind are that addition and multiplication of ordinals is associative, $n\cdot \omega = \omega$ for any $n \in \mathbb{N^{+}}$, and $\alpha \cdot 2 = \alpha + \alpha$. We also have a left distributive law, $\alpha\cdot (\beta + \gamma) = \alpha \cdot \beta + \alpha \cdot \gamma$, but not a right distributive law.

For (i), we have $(\omega + 1)\cdot 2 = (\omega + 1) + (\omega + 1) = ((\omega + 1) + \omega) + 1$ $$ = (\omega + (1 + \omega)) + 1 = (\omega + \omega) + 1 = \omega\cdot 2 + 1$$

On the other hand $2\cdot (\omega + 1) = 2\cdot \omega + 2 = \omega + 2$

Hopefully it is clear that $\omega + 2 < \omega\cdot 2 + 1$. Therefore $(\omega + 1)\cdot 2 > 2\cdot (\omega + 1)$.

For (ii), $$(\omega + 1)\cdot \omega = \mbox{sup}_{n < \omega} (\omega + 1)\cdot n$$ $$= \omega^2 + 1$$

On the other hand $$\omega\cdot (\omega + 1) = \omega^2 + \omega$$

Hopefully it's clear that $\omega^2 + 1 < \omega^2 + \omega$. Thus $(\omega + 1)\cdot \omega < \omega \cdot(\omega + 1)$