**Background**


The fact that there is no suborder of $\mathbb R$ which is of type $\omega\_1$ suggests (to me) that the continuum $c$ cannot be very far from $\omega\_1$: How could $c$ be far away from $\omega\_1$ if there is no room for an order embedding of $\omega\_1$ in $\mathbb R$? Of course, this fact is a consequence of the separability of $\mathbb R$ (which is itself an amazing fact: How can continuum many aligned irrationals be separated by only countably many rationals?) 


From the idea that the continuum cannot be very far from $\omega\_1$ because there is no room in $\mathbb R$ to embed $\omega\_1$, one can easily formulate an axiom implying $CH$: 


**Preliminary definition**


Let $\kappa$ be an infinite cardinal and $L$ be a total order. We say that $L$ is $\kappa$-unbounded if $|L|=\kappa$ and for every $a\in L$, we have that
$|\left\{x\in L : a < x\right\}|=\kappa$.


**The axiom**


If $\kappa$ and $\lambda$ are infinite cardinals, $\lambda<\kappa$, and $L$ is a $\kappa$-unbounded total order, then there is an order embedding $f: \lambda\rightarrow L$ (in other words, there is a suborder of $L$ of type $\lambda$).


I have not double-checked every detail, but I am convinced that this axiom easily implies $GCH$ for at least all strong limit cardinals. In particular, it implies $CH$.


**Question**


Is the above axiom consistent with $ZFC$? Maybe there is an easy counterexample, but I have not found one. I know that this is related to the dense set problem as presented in Baumgartner, J., *Almost disjoint sets, the dense set problem and the partition calculus*.


**EDIT**


In view of Goldstern's counterexample, here is a modification of the axiom that might be consistent with ZFC:


Assume that $\kappa$ and $\lambda$ are infinite cardinals, $\lambda<\kappa$, $L$ is a $\kappa$-unbounded total order and that $L^\*$ (the reverse order) is also $\kappa$-unbounded. Then there is an order embedding $f: \lambda\rightarrow L$ or an order embedding $g: \lambda^\*\rightarrow L$ (in other words, there is a suborder of $L$ of type $\lambda$ or of type $\lambda^\*$).


This still implies $CH$ and the given counterexamples do not apply. 



Consistency of embedding cardinals in linear orderings
Your axiom is inconsistent. (Or perhaps I have misunderstood it.)


Let $L\_n:= \aleph\_n$ with the reverse order, and let $L:= L\_1 + L\_2 + \cdots$ (horizontal sum); equivalently, let $L$ be the lexicographic order on $\bigcup\_k \{k\}\times L\_k$. Then $L$ is $\aleph\_\omega$-unbounded, yet there is no order-preserving embedding of $\omega\_1$ into $L$: (EDITED to simplify:) Every well-ordered subset of $L$ is finite in each $L\_n$, hence at most countable.


