Unions of ordinals and their suprema

Question

I am trying to construct a nested collection $\{S_k:\,k\in \omega\}$ of sets of ordinals such that the order type of $\bigcup_k S_k$ is not the supremum of the order types of $S_k$, and to also give a condition on the $S_k$ such that no such construction is possible (as worded in the problem).

Inspired by $1+2+\cdots+\omega=\omega\cdot 2$ I figured enumerating $\{\alpha:\;\alpha<\omega+1\}=\{S_0,S_1\dots\}$ and taking $\bigcup S_i$ would give me $\omega\cdot 2$, where as the sup of the order types is just $\omega$ (or $\omega+1$, not sure). Is that correct?

My initial plan was actually $S_k=\{0,\dots,k,\omega\}$ (since the question just stipulates "sets of ordinals"), but then I realised we might just have $\bigcup S_k=\omega$ and in fact order type of $S_k\neq\omega$ but possibly just $k+1$. Confused about the whole idea basically, a comment on that would be helpful too.

Last, as for the condition, at first I thought if the $S_k$ are eventually a chain of successors then this union will be the $\sup$ of the order types but again I'm not sure whether this is the condition we are after or whether it is even true.

I would really appreciate a little help with understanding the concepts at play here...

Answer 1

Your initial plan works out just fine. Each $S_k$ has order type $k+1$, so the supremum of the order types is $\omega$. But the union is $\omega+1$ which is strictly larger.

The reason that the union is $\omega+1$ is that $\omega$ is a member of the union, and the union is in fact an ordinal, so it cannot be $\omega$ itself. You can also play with this to obtain any ordinal below $\omega\cdot2$; but also $\omega\cdot\omega$ by taking $S_k=\{0,\ldots,k,\omega,\ldots,\omega+k\}$.

Essentially, whenever $\alpha$ is a countable ordinal, you can enumerate it, and define a sequence of finite sets which increase towards $\alpha$; then if $\alpha>\omega$ you get something like that. Note that as $S_k$ is not necessarily finite, you can use this to conclude something on the cofinality of the ordinals at play here.

Answer 2

Ordinal exponentiation: For ordinals $A,B$ the ordinal $A^B$ is the order-type of the set $S(A,B)$ of functions $f:B\to A$ such that $\omega > |\{b\in B:f(b)\ne 0\}|,$ with the reverse-lex order $<_{r}$: When $f,g \in S(A,B)$ with $f\ne g$ and $c=\max \{b\in B:f(b)\ne g(b)\}$ then $f<_{r}g\iff f(c)<g(c).$

Let $A=\omega$ and $B=\omega +1.$ For $n\in \omega,$ let $$S_n=\{f\in S(A,B): f(\omega)\leq n \land \forall x\in \omega \;(f(x)\ne 0\implies x\leq n)\}.$$ Then $S_n\subset S_{n+1}$ and OT$(S_n)=\omega^{n+1}\cdot (n+1)<\omega^{n+1}\cdot \omega=\omega^{n+2}<\omega^{\omega}.$ So we have $$\sup_{n\in \omega}\text {OT}(S_n)=\omega^{\omega}< \omega^{\omega +1}=\cup_{n\in \omega}S_n=\text {OT}(\cup_{n\in \omega}S_n).$$

This is a variation on the theme of "The Rado-Milner Paradox": If $k$ is an infinite cardinal and $k\leq x<k^+$ there exists $T=\{K_n:n\in \omega\}$ with $\cup T=x,$ and OT$(K_n)\leq k^n$ for each $n\in \omega.$(Where $k^n$ denotes ordinal exponentiation.)