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If $\kappa=\kappa^{<\kappa}$ then there exists a dense linear order of size $\kappa$ in which every ordinal of cardinality $\leq\kappa$ can be embedded.

I saw somewhere the set I am looking for is $L$ the functions from $\kappa \rightarrow \kappa$ with bounded support.

The proof would seem to be the following: by induction the empty set and successor cases are relatively easy. For the limit case I can use the concatenation function $\{\alpha\}^\frown:L\rightarrow L$ which sends everything in a sub interval. So given $\delta$ a limit ordinal less than $\kappa$ there is a $\lambda\leq\kappa$ sequence $(\alpha_\beta)_{\beta<\lambda}$ cofinal in $\delta$ for each $[\alpha_\beta,\alpha_{\beta+1})$ embed in $L$ with $\varphi_\beta$. Compose each $\varphi_\beta$ and compose it with the concatenation with $\alpha$. then taking the union of all these functions gives the needed embedding of $\delta$ in $L$. This seems to work with functions from $\omega\rightarrow\kappa$ that have bounded support.

I would appreciate any clarification of this fact and whether there is an error in the proof and how it would fail when considering the set of $\omega $ sequences.

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  • $\begingroup$ The idea of your proof seems correct, but you may simply the proof: we can see that $L$ and $L_\beta=\{f\in L \mid f(0)=\beta\}$ are isomorphic, $L_\beta<L_\gamma$ if $\beta<\gamma$, and we can embed each $\alpha_{\beta+1}\setminus \alpha_\beta$ into $L_\beta$. $\endgroup$
    – Hanul Jeon
    Aug 12, 2021 at 9:54
  • $\begingroup$ Just every ordinal, not every linearly ordered set of cardinality $\le\kappa$, can be embedded? $\endgroup$
    – bof
    Aug 14, 2021 at 0:55
  • $\begingroup$ @bof yes just ordinals $\endgroup$ Aug 17, 2021 at 12:37

2 Answers 2

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I'd like to give a bit of model-theoretic context for your question, which is in a certain sense a special case of a more general phenomenon in model theory. This is not an answer to your question, but it's slightly too long for a comment, so I hope you'll excuse me for writing it here. Let $T$ be any first-order theory, and suppose that $\kappa$ is a cardinal such that $\kappa>|T|$ and $\kappa=\kappa^{<\kappa}$. Then $T$ has a saturated model $M$ of size $\kappa$. (Let me know if you would like a reference to this fact; the idea is to build an elementary chain of length $\kappa$, each link of which realizes types over the earlier links.) Now, for a cardinal $\lambda$ and a structure $M$, we say that $M$ is "$\lambda$-universal" if every elementarily equivalent structure $N\equiv M$ of size $<\lambda$ admits an elementary embedding into $M$. One can show that a saturated model $M$ is $|M|^+$-universal; again let me know if you would like a reference to this fact.

So, putting these two facts together, we get the following fact for any cardinal $\kappa$ with $\kappa=\kappa^{<\kappa}$:

If $T$ is a complete first-order theory with $|T|<\kappa$, then there exists a model $M\models T$ of cardinality $\kappa$ such that any model of $T$ of cardinality $\leqslant\kappa$ is isomorphic to an elementary substructure of $M$.

Taking the case where $T=\text{DLO}$ is the theory of dense linear orders gives your desired result almost immediately; indeed, we then get a dense linear order $M$ of size $\kappa$ such that every dense linear order of size $\leqslant\kappa$ embeds elementarily into $M$. If $(O,<)$ is now any linear order of size $\leqslant\kappa$, then $O\times\mathbb{Q}$ with lexicographic ordering is a dense linear order of size $\leqslant\kappa$, and hence there exists an elementary embedding $O\times\mathbb{Q}\to M$. Taking the composition of this with the embedding $O\to O\times\mathbb{Q}$ given by $a\mapsto (a,0)$ now gives an order-embedding of $O$ into $M$. Thus any linear order of size $\leqslant\kappa$ has an order-embeding into $M$, and so in particular every ordinal of size $\leqslant\kappa$ does too.


Of course, this is a very "non-constructive" proof, and it does not give you any tractable representation of $M$ to work with; for some theories $T$, it will generally not be possible to give an explicit construction of the model $M$. $\text{DLO}$ is thus a nicer theory in this regard, and the proof you are giving in your post is a much nicer and more concrete solution for that case. But I thought it might be worthwhile to point out that, provided such a cardinal $\kappa$ exists, then every first-order theory (of size $<\kappa$) will have a model of size $\kappa$ with analogous properties.

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    $\begingroup$ thank you for the response. My doubt was that it appears I can repeat the proof for finite sequences of ordinals bellow $\kappa$ it would seem the proof would still hold. This would mean that for all cardinals there is a DLO in which ordinals of the same cardinality embed. $\endgroup$ Aug 17, 2021 at 12:40
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    $\begingroup$ I found out the answer to my question but I will leave you the bounty. If you could send me a reference to that theorem I would appreciate it. $\endgroup$ Aug 19, 2021 at 14:50
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    $\begingroup$ dear @aldodecristo oh dear, apologies for missing your comment a few days, I did not see it! but I am very glad that you have nonetheless answered your question. (+1 on your answer.) thank you so much for the bounty, I do not know that my answer deserves it!! anyway, regarding references, check for example Theorem I-1.7 in Shelah's Classification Theory. but I would like to make a note that one can actually carry out similar constructions in ZFC alone; in particular, given any infinite cardinal $\mu$, define $\kappa=\beth_{\mu^+}(\mu)$ ... $\endgroup$ Aug 19, 2021 at 22:21
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    $\begingroup$ @aldodecristo ... then, in ZFC, one can prove that any structure $M$ of size $\leqslant\kappa$ has what is called a "special" elementary extension of size $\kappa$. special structures have the property that they are $\kappa^+$-universal, so this gives a way of carrying out a similar construction even without the assumption that there exists some $\lambda$ with $\lambda=\lambda^{<\lambda}$. for details on this construction, see Section 6.1 of Tent and Ziegler's A Course in Model Theory. I hope this helps! :) and thank you once again for the bounty $\endgroup$ Aug 19, 2021 at 22:22
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The reason why I asked this question is that $L$ is used to build an $\kappa^+$ Aronszajn tree. In Specker's articles "Sur un probléme de Sikorski" he provides a detailed proof of how one can construct such a tree from an order $L$ (or $E_\nu$ in his case) with the following properties:

(i) is dense
(ii) has cardinality $\kappa$
(iii) any ordinal $<\kappa^+$ can be embedded in it

but also:

(iv) that given $\alpha<\kappa$ and $s$ an $\alpha$ increasing sequence we have that if $a>s(\xi)$ for all $\xi<\alpha$ then there is $a>b>s(\xi)$. In other words no $\alpha<\kappa$ sequence can have a supremum that is not a maximum.

(i),(ii), and (iii) hold for the set of finite sequences and so for any infinite cardinal there is a linear order with such properties. For (iv) to hold we need to use the fact that $\kappa=\kappa^{<\kappa}$ and we need to slightly modify $L$ to be functions from $\lambda\rightarrow \lambda_\pm$ where by $\lambda_\pm$ we mean $\lambda$ with the corresponding negative ordinals added.

Given an $\alpha<\kappa$ sequence $s$ in $L$ and $a$ an upper bound we have that $\bigcup_{\xi<\alpha}supp(s(\xi))\subseteq \beta<\kappa$ must be bounded since $\kappa$ must be regular (were it singular $\kappa^{<\kappa}\geq \kappa^{cf(\kappa)}>\kappa$). We take $\gamma>\beta$ to be such that $supp(a)\subseteq \gamma$. Now taking $b=a$ except on the $\gamma$ coordinate where we set $b(\gamma)=-1$. We have that $b$ is still an upper bound to $s$ but $b<a$.

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