Let us say that a mathematical structure of cardinality $\omega\_1$ is *Jonsson* whenever every one of its proper substructures is countable.


There are examples of Jonsson groups due to [Shelah](http://www.sciencedirect.com/science/article/pii/S0049237X08713466) or Obratzsov. I am almost sure that there is no Jonsson Boolean algebra but I cannot (dis)prove it by hand. Am I right? 


PS. feel free to give any further examples of Jonsson structures or structures which are never Jonsson.



Jonsson Boolean algebras?
Boolean algebras are never Jonsson. 


Suppose that $\mathbb{B}$ is a Boolean algebra of size $\omega\_1$. Let $a$ be any element such that neither $a$ nor $\neg a$ is an atom. Note that every element $b\in\mathbb{B}$ is the join $b=(b\wedge a)\vee(b\wedge \neg a)$, and so there must be uncountably many elements either in the cone below $a$ or below $\neg a$. Assume without loss of generality that there are uncountably many elements below $a$. Let $\mathbb{C}$ be the subalgebra of $\mathbb{B}$ consisting of the elements below-or-equal $a$ or above-or-equal $\neg a$. This is closed under meets, joins and complements, and hence is a sub-Boolean algebra. And it has size $\omega\_1$ by the choice of $a$. But it has no elements below $\neg a$ other than $0$, and so $\mathbb{C}$ is an uncountable proper subalgebra, as desired. QED 


It seems that the same idea generalizes to any uncountable cardinal.


