Sum of finite ordinals: $\lambda_1+\lambda_2+\dots+\lambda_n+\dots=\omega$

Question

Prove: $\lambda_1+\lambda_2+\dots+\lambda_n+\dots=\omega$, where $\lambda_i$ are finite ordinal nonzero numbers.

I tried like this.

$A_i$ set and ord($A_i$)$=\lambda_i, i=\{1,2,...\}$ and ord($\mathbb{N}$)$=\omega$.

$\lambda_1+\lambda_2+\dots+\lambda_n+\dots=$ord($\bigcup (A_i$x$\{i\})$)

$\bigcup (A_i$x$\{i\})$ is countable union of finite set. And that is countable. So ord($\bigcup (A_i$x$\{i\})$)=ord($\mathbb{N}$)

Is this correct?

EDIT: Definition: $\{\lambda_i\}_{i\in I}$ is family of ordinal numbers and $\lambda_i=$ord($A_i$). Then $\sum_{i\in I}=$ord($\bigcup_{i\in I} (A_i$x$\{i\})$)

I think that I need to find order isomorphism from $\bigcup_{i\in I} (A_i$x$\{i\})$ to $\mathbb{N}$ but I'm not sure how.

Given two posets $(S,\le_S)$ and $(T,\le_T)$, and order isomorphism from $(S,\le_S)$ to $(T,\le_T)$ is bijective function $f$ from $S$ to $T$ with the property that, for every $x,y\in S$, $x\le_S y$ iff $f(x)\le_T f(y)$

Answer 1

It is easy to show the sum is not finite. The sum is a limit of finite sums however. the only candidate is $\omega$.

Answer 2

Note $\lambda_{i} = 1 + 1 + 1 \cdots$ ($\lambda_{i}$ times). Replace $\lambda_{i}$ accordingly in the sum. By the associative property we can remove parentheses and get $(1 + 1 + 1 + \cdots)$ which is obviously $\omega$.

Answer 3

You are correct in thinking that getting an order isomorphism between

$$A=\bigcup_{k\in\Bbb Z^+}\big(A_k\times\{k\}\big)$$

and $\Bbb N$ would prove the desired result. We have a natural linear order $\preceq$ on $A$, though it’s a little messy to describe. For each $k\in\Bbb Z^+$ you have an order isomorphism $h_k:A_k\to\lambda_k$. Given $\langle a,k\rangle,\langle b,\ell\rangle\in A$, we set $\langle a,k\rangle\preceq\langle b,\ell\rangle$ if and only if either $k<\ell$, or $k=\ell$ and $h_k(a)\le h_k(b)$. This has the effect of putting $A_1\times\{1\}$ first, ordered in the order corresponding to $h_1$, followed by $A_2\times\{2\}$ in the order corresponding to $h_2$, and so on.

It’s pretty clear intuitively that each element of $A$ has only finitely many predecessors in the order $\preceq$, so we ought to be able to get an order isomorphism $h:A\to\Bbb N$ by mapping each $a\in A$ to the number of predecessors that it has in the linear order $\langle A,\preceq\rangle$. (Note: My $\Bbb N$ includes $0$.)

Consider an arbitrary $\langle a,k\rangle\in A$; how many predecessors does it have? Every $\langle b,\ell\rangle$ with $1\le\ell<k$ and $b\in A_\ell\times\{\ell\}$ is a predecessor and there are $\sum_{\ell=1}^{k-1}\lambda_\ell$ of them. (If $k=1$ the sum has no terms and is therefore $0$.) Moreover, $a$ has $h_k(a)$ predecessors in $A_k$, so altogether we want to set

$$h\big(\langle a,k\rangle\big)=h_k(a)+\sum_{\ell=1}^{k-1}\lambda_\ell\;.\tag{1}$$

This is a sum of finitely many non-negative integers, so it’s in $\Bbb N$. It’s straightforward to check that the function $h$ defined by $(1)$ is strictly order-preserving, so all that remains is to show that it’s surjective.

Let $s_0=0$, and for $n\in\Bbb Z^+$ let $s_n=\sum_{k=1}^n\lambda_k$; clearly $s_n\ge n$. Let $n\in\Bbb N$; it follows that there is a unique $m\in\Bbb Z^+$ such that $s_{m-1}\le n<s_m$. Now

$$n-s_{m-1}<s_m-s_{m-1}=\lambda_m\;,$$

so there is a unique $a\in A_m$ such that $h_m(a)=n-s_{m-1}$. But then

$$h\big(\langle a,m\rangle\big)=h_m(a)+\sum_{\ell=1}^{m-1}\lambda_\ell=n-s_{m-1}+s_{m-1}=n\;,$$

and $h$ is indeed surjective and hence the desired order isomorphism.