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I am having trouble with some simple ordinal arithmetic.
I am trying to figure out what $(\omega+2)\cdot \omega$ is.
If you represent $\omega + 2$ as: $\omega+2 = 0_0 < 1_0 < 2_0 < ... < \omega + 1 < \omega + 2$
Then when I try to solve it out by writing out the representations, I get:
$ \begin{align} &0_0 < 1_0 < 2_0 < ... < (\omega+1)_0 < (\omega+2)_0 \\ < &0_1 < 1_1 < 2_1 < ... < (\omega+1)_1 < (\omega+2)_1 \\ &... \end{align} $
Which if I just rename the terms becomes:
$ \begin{align} &0_0 < 1_0 < 2_0 < ... \\ < &0_1 < 1_1 < 2_1 < ... \\ &... \end{align} $
Or simply: $\omega + \omega + \omega + ... = \omega^2$

Is this right? or am I thinking about this the wrong way

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You’re right: $(\omega+2)+\omega=\omega+\omega$, so the extra $+2$’s get absorbed in the following $\omega$. You can prove by induction that $(\omega+2)\cdot n=\omega\cdot n+2$, so you have $$\omega\cdot n<\omega\cdot n+2<\omega\cdot(n+1)$$ for each $n\in\omega$, and hence the supremum of the $\omega\cdot n+2$’s is just the supremum of the $\omega\cdot n$’s, or $\omega^\omega$. – Brian M. Scott Jan 16 '13 at 20:12
Yeah, just don't fall for fallacies like "All the intermediate 2's get swallowed, but the last remains, so it's $\omega^2+2$". – Hagen von Eitzen Jan 16 '13 at 20:35
Hagen, now that you mention this, why is that a fallacy? – sicklybeans Jan 16 '13 at 20:58
Ah I figured it out – sicklybeans Jan 16 '13 at 21:25

Personally I do like the inductive definition for working these sort of things out, when the expression is simple enough it can be quite illuminating too:

Recall that $\alpha\cdot\omega=\sup\{\alpha\cdot n\mid n\in\omega\}$. In this case we have, if so:

$$(\omega+2)\cdot\omega=\sup\{(\omega+2)\cdot n\mid n\in\omega\}=\sup\{(\omega+2)\cdot n\mid n\in\omega\}$$

But what is $(\omega+2)\cdot n$? Well, recall that $\alpha\cdot(\beta+\gamma)=\alpha\cdot\beta+\alpha\cdot\gamma$. Therefore $$(\omega+2)\cdot n=\underbrace{\omega+2+\omega+2+\ldots+\omega+2}_{n\text{ times}}=\omega\cdot n+2$$

So what we want to know is what is $\sup\{\omega\cdot n+2\mid n\in\omega\}$, but this is not a difficult exercise to see that this is indeed $\omega^2$.

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Thank you, this was helpful. Now I have another question:… – sicklybeans Jan 16 '13 at 21:31

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