Ordinal arithmetic So I'm having trouble understanding ordinal arithmetic. 
So if you have $\omega = \bigcup \{ n | n\in\mathbb{N}\}$
How is this defined 
$\omega^2$ as in the notes I'm reading it has $\omega^2 = \bigcup \{ \omega^n| n<\omega\}$, however this can't be correct. 
So I'm having big trouble of understanding multiplication.
I understand how addition works. 
In particular, if $\eta$ is the order type of $\mathbb{Q}$, what is $(\eta+1) \cdot (\eta+1)$?
 A: I suspect that the definition of $\omega^2$ was intended to read
$$\omega^2=\bigcup\{\omega\cdot n:n<\omega\}\;,$$
where $\cdot$ indicates ordinal multiplication. Using $\longrightarrow$ to represent a set ordered in type $\omega$, we can picture $\omega\cdot n$ as
$$\underbrace{\longrightarrow\longrightarrow\ldots\longrightarrow}_{n\text{ arrows}}$$
and get the following picture:
$$\begin{array}{rl}
\omega\cdot 0:\\
\omega\cdot 1:&\longrightarrow\\
\omega\cdot 2:&\longrightarrow\longrightarrow\\
\omega\cdot 3:&\longrightarrow\longrightarrow\longrightarrow\\
\vdots\quad&\qquad\vdots\\
\omega^2=\omega\cdot\omega:&\underbrace{\longrightarrow\longrightarrow\longrightarrow\longrightarrow\longrightarrow\ldots}_{\omega\text{ arrows}}
\end{array}$$

The order type $\eta+1$ is what you get when you add a right endpoint to $\Bbb Q$; it’s the order type of $\Bbb Q\cap(0,1]$, for instance. The order type $(\eta+1)\cdot(\eta+1)$ can then be visualized as follows.

Let $A=\Bbb Q\cap(0,1]$, and let $X=A\times A$. Let $\le$ be the usual linear order on $A$, and let $\preceq$ be the reverse lexicographic order on $X$ defined as follows: if $\langle p,q\rangle,\langle a,b\rangle\in X$, $\langle p,q\rangle\preceq\langle a,b\rangle$ iff either $q<b$, or $q=b$ and $p\le a$.

$X$ is the set of points of $(0,1]\times(0,1]$ with rational coordinates. Speaking pictorially, if $x$ and $y$ are points of $X$ with $x$ to the left of $y$, then $x\preceq y$; if $x$ and $y$ are on the same vertical line, then $x\prec y$ iff $x$ is below $y$.
A: Another way of visualizing $\omega^2$ as an order type is to embed it into the rationals: since the numbers $\left\{0, \frac{1}{2},\frac{3}{4}, \frac{7}{8},\ldots\right\}$ are an embedding of $\omega$ into the interval $\left[0,1\right)$, we can find our embedding of $\omega^2$ by simply placing $\omega$ copies of these — one for each natural number — side by side: $\left\{0, \frac{1}{2}, \frac{3}{4},\frac{7}{8},\ldots,1,1\frac{1}{2}, 1\frac{3}{4}, \ldots, 2, 2\frac{1}{2}, 2\frac{3}{4}, \ldots, 3, \ldots \right\}$ - in other words, all the numbers of the form $i+1-2^{-j}$, $i,j\geq 0$.
