If $\kappa=\kappa^{<\kappa}$ then there is a dense linear order of size $\kappa$ in which all ordinals $<\kappa^+$ can embed.

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.

• 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 $L_\beta$. Aug 12, 2021 at 9:54
• Just every ordinal, not every linearly ordered set of cardinality $\le\kappa$, can be embedded?
– bof
Aug 14, 2021 at 0:55
• @bof yes just ordinals Aug 17, 2021 at 12:37

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.

• 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. Aug 17, 2021 at 12:40
• 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. Aug 19, 2021 at 14:50
• 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 $\kappa=\beth_{\mu^+}(\mu)$ ... Aug 19, 2021 at 22:21
• @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 Aug 19, 2021 at 22:22

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.