Short question about ordinals multiplication definition from wiki Why does $2\cdot\omega$ looks like this: $0_0 < 1_0 <0_1 <1_1 ...$ ? Is that another way to represent $1_0 < 1_1 < 2_0 <2_1 <3_0<3_1 ...$ ?
Edit: also, why does it equal to omega ?
Link: http://en.wikipedia.org/wiki/Ordinal_arithmetic
 A: Ordinal multiplication is defined by induction of the right multiplicand as follows:


*

*$\alpha \cdot 0 = 0$;

*$\alpha \cdot ( \beta + 1 ) = ( \alpha \cdot \beta ) + \alpha$;

*$\alpha \cdot \lambda = \lim_{\beta < \lambda} \alpha \cdot \beta$ for limit $\lambda > 0$.


So $2 \cdot \omega = \lim_{n < \omega} 2 \cdot n$.  Let's look at the first few levels of this:


*

*well, $2 \cdot 0 = 0$, nothing to see here.

*$2 \cdot 1 = ( 2 \cdot 0 ) + 2 = 0 + 2 = 2$;

*$2 \cdot 2 = ( 2 \cdot 1 ) + 2 = 2 + 2$;

*$2 \cdot 3 = ( 2 \cdot 2 ) + 2 = ( 2 + 2 ) + 2 = 2 + 2 + 2$;

*$2 \cdot 4 = ( 2 \cdot 3 ) + 2 = ( 2 + 2 + 2 ) + 2 = 2 + 2 + 2 + 2$


In general, $2 \cdot n$ is $n$ copies of "$2$" (meaning a two-element well-ordered set) placed back-to-back, so in the limit we get $\omega$ copies of "$2$" placed back-to-back.  What representation you give these copies is pretty much arbitrary, so writing $$0_0 < 1_0 < 0_1 < 1_1 < 0_2 < 1_2 < \cdots\tag{1}$$ doesn't really differ much from $$1_0 < 1_1 < 2_0 < 2_1 < 3_0 < 3_1 < \cdots\tag{2}$$  (you can translate from from (1) to (2) via the mapping $i_j \mapsto (j+1)_i$).  This first representation may be somewhat pedagogically preferred, since $0 < 1$ is the prototypical 2-element well-ordered set, and the subscripts then indicate which copy of this order you are in.
If you think about the limit, it is not too difficult to see that it is $\omega$.  First note that $m < n < \omega$ implies that $2 \cdot m < 2 \cdot n$, and so $\lim_{n < \omega} 2 \cdot n$ is simply the supremum of the set $\{ 2 \cdot n : n < \omega \}$. (This is generally true: ordinal multiplication is always non-decreasing in the right multiplicand, so given any $\alpha$ and limit $\lambda > 0$ the product $\alpha \cdot \lambda$ is the supremum of $\{ \alpha \cdot \beta : \beta < \lambda \}$.  Similarly for the other usual ordinal operations.)


*

*It cannot be greater than $\omega$ since $2 \cdot n$ is finite for all $n < \omega$ (and so $\omega$ is an upper bound of the set $\{ 2 \cdot n : n < \omega \}$).

*It cannot be less than $\omega$ since $n < n+1 < 2 \cdot ( n+1 ) \leq 2 \cdot \omega$.

