Power two of ordinal

Question

if $\omega$ is ordinal number of set $\mathbb{N}$ can I compute $(\omega + 1)^2$ like below

$\begin{align*} (\omega + 1)^2 & = \ (\omega + 1).(\omega + 1)\\ & = \ ((\omega + 1).\omega)+(\omega + 1)\\ & = \ (\omega^2 + 1)+(\omega + 1)\\ & = \ \omega ^ 2 + (1 + \omega) + 1\\ & = \ \omega^2 + \omega + 1 \end{align*}$

is $(\omega+1).\omega = \omega^2 + 1$? how can I compute $(\omega+1).\omega$?

Answer 1

Your answer is right, but your argument is wrong.

Remember that $\alpha\cdot \beta$ means "$\beta$ many copies of $\alpha$" - so e.g. $\omega\cdot 2=\omega+\omega$, whereas $2\cdot\omega=2+2+2+2+...=\omega$.

With this in mind, we have $$(\omega+1)\cdot\omega=\omega+1+\omega+1+\omega+1+...=\omega+(1+\omega)+(1+\omega)+ . . . =\omega^2.$$

So indeed we have $(\omega+1)\cdot(\omega+1)=\omega^2+\omega+1$.

However, it looks like in your argument you assume (third line) that $(\omega+1)\cdot\omega=\omega^2+1$, which is not correct.

Answer 2

Sometimes a picture can help, so I’m adding this addendum to Noah’s answer for the possible benefit of future readers. The first diagram shows the Cartesian product $\omega\times(\omega+1)$; $(\omega+1)\cdot\omega$ is the order type of this product under the lexicographic order.

$$\begin{array}{ccc} \omega&\color{red}\omega&\color{blue}\omega&\color{green}\omega&\ldots\\ \vdots&\vdots&\vdots&\vdots&\\ 4&4&4&4&\ldots\\ 3&3&3&3&\ldots\\ 2&2&2&2&\ldots\\ 1&1&1&1&\ldots\\ 0&0&0&0&\ldots \end{array}$$

Now imagine taking each of the points in the top row and moving it to the bottom of the next column to the right; the result is shown below.

$$\begin{array}{ccc} \vdots&\vdots&\vdots&\vdots&\vdots\\ 5&4&4&4&4&\ldots\\ 4&3&3&3&3&\ldots\\ 3&2&2&2&2&\ldots\\ 2&1&1&1&1&\ldots\\ 1&0&0&0&0&\ldots\\ 0&\omega&\color{red}\omega&\color{blue}\omega&\color{green}\omega&\ldots\\ \end{array}$$

The lexicographic order on this array is clearly isomorphic to that on the first array, but here it clearly has order type $\omega\cdot\omega=\omega^2$.