**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.

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