Why is $n+\omega=\omega$ for finite $n$?

Question

I am trying to understand the following claim;

$n+\omega=\omega$ where $n$ is order type of a finite set and $\omega$ is the order type of $\left\{ 1,2,\dots, \right\}$ with the usual meaning of $<$.

My question is how is it possible, for instance, $\left\{1,2,\dots ,n,1,2,\dots \right\}$ has order $\omega$ whilst the order $\left\{1,2,\dots\right\}$ is $\omega$? Many thanks for any help.

Answer 1

Here’s a quick sketch of the matchup:

    1   2   3   4  ...  n  n+1 n+2 n+3 n+4 n+5 n+6 ...
    1   2   3   4  ...  n   1   2   3   4   5   6  ...  

The actual function mapping the top line to the bottom line is:

$$f:\Bbb Z^+\to\Bbb Z^+:f(k)\mapsto\begin{cases} k,&\text{if }k\le n\\ n-k,&\text{if }k>n\;. \end{cases}$$

You almost had it in your comment; you just didn’t make sure that you matched up the two sets completely.

Answer 2

There are two possible ways to define ordinal addition:

1. We define $\alpha+\beta$ to be the unique $\gamma$ which is order isomorphic to $\{0\}\times\alpha\cup\{1\}\times\beta$ with the lexicographic order.
2. We define the addition by recursion from the successor function. $\alpha+0=\alpha$; $\alpha+(\beta+1)=(\alpha+\beta)+1$; and for limit $\delta$, $\alpha+\delta=\sup\{\alpha+\beta\mid\beta<\delta\}$.

Given the first definition, you need to find an order isomorphism between $\omega$ and $\{0\}\times n\cup\{1\}\times\omega$. Luckily, between well-ordered sets there exists only one isomorphism, so if you found one you found them all. The is to push the elements of $\omega$ over the first $n$ places, and then put $n$ in there. You can work out the details on your own.

Given the second definition, however, things are much much easier. $$n+\omega=\sup\{n+m\mid m<\omega\}=\omega.$$