The starting point of this question is the observation that if $\lambda$ is a countable ordinal, then there is an order-embedding $e:\lambda \hookrightarrow \mathbb{Q}$.


Given an infinite cardinal $\kappa$, is there a linearly ordered set $L\_\kappa$ of cardinality $\kappa$ such that for every ordinal $\alpha$ with $|\alpha|=\kappa$ there is an embedding $e:\alpha \hookrightarrow L\_\kappa$?



Ordinal-universal linear order on $\kappa$ elements
Let $\lambda$ be an infinite cardinal. Give $[\lambda]^{<\omega}$ the linear ordering $\leq$ where we set $A>B$ if there is some $\alpha\in B$ where $\alpha\cap A=\alpha\cap B$ and $\alpha\not\in A$.


We shall show using what is essentially transfinite induction that every ordinal less than $\lambda^+$ embeds into $[\lambda]^{<\omega}$. Suppose that $\alpha$ is the least ordinal where $\alpha$ does not embed into $\lambda^+$. Clearly $\alpha>0$. If $\alpha$ is a successor ordinal, then $\alpha=\beta+1$ for some $\beta$ and there is some embedding $f:\beta\rightarrow[\lambda]^\omega$. Now, define $g:\alpha\rightarrow[\lambda]^\omega$ by letting
$g(\delta)=\{0\}\cup\{1+\sigma\mid \sigma\in f(\delta)\}$ whenever $\delta<\alpha$ and by letting $g(\beta)=\emptyset$. Then $g$ is an order preserving embedding, so $\alpha$ cannot embed into $\lambda^+$. Now suppose that $\alpha$ is a limit ordinal. Then let $(x\_\beta)\_{\beta<\gamma}$ be a cofinal continuous increasing sequence in $\alpha$ with $x\_0=0$ where $\gamma$ is a cardinal with $\gamma\leq\lambda$. Then for each $\beta<\gamma$, let
$f\_\beta:[x\_\beta,x\_{\beta+1})\rightarrow[\lambda]^{<\omega}$ be an embedding. Then define an embedding $f:\alpha\rightarrow[\lambda]^{<\omega}$ by letting
$f(x)=\{\beta\}\cup\{\beta+1+\delta\mid \delta\in f\_\beta(x)\}$ whenever $x\in[x\_\beta,x\_{\beta+1})$.


Observe that $[\lambda]^n$ is a well-ordered of order type $\lambda^n$ whenever $n\in\omega$. Therefore $[\lambda]^{<\omega}$ is the union of countably many well-ordered sets. Every totally ordered set $X$ which is the union of countably many well-ordered sets has no descending sequence of length $\omega\_1$. Therefore, $[\lambda]^{<\omega}$ has no descending sequence of length $\omega\_1$.


