Is not SH + not CH consistent? I guess that $\lnot$SH + $\lnot$CH is consistent, but I have not found this question discussed anywhere. Is there any relatively simple model of $\lnot$SH + $\lnot$CH?
 A: Recall that adding a single Cohen real adds a Suslin tree, and does not change the value of the continuum [1] [2]. So by taking any model of $\lnot\sf CH$ and adding a single Cohen real, we have a Suslin tree, and therefore $\lnot\sf SH$ as well.
(This makes the Solovay-Tennenbaum theorem all the more magical, since we add so many Suslin trees in the iteration, but we still manage to kill them all eventually...)



*

*Jech, Set Theory, 3rd Millennium ed., Theorem 28.12, p.563

*Saharon Shelah, Can you take Solovay’s inaccessible away?, Israel J. Math. 48 (1984), no. 1, 1--47.
A: Two more solutions. 


*

*Start with a model with a Suslin tree $(T,\leq)$ in it 
(force it or use L). Let $(P,\leq)$ be the Cohen forcing adding 
any number of reals. We claim that $(T,\leq)$ remains Suslin after 
forcing with $(P,\leq)$. Assume that $p$ forces that $A$ is an 
uncountable antichain in $T$. There are conditions $p_\alpha\leq p$ 
and distinct elements $t_\alpha\in T$ such that $p_\alpha$ forces 
that $t_\alpha\in A$. An easy argument (using the delta system lemma) 
shows that there is an uncountable set $Z$ such that 
any finite subfamily of $\{p_\alpha:\alpha\in Z\}$ has a common lower bound. 
Then $\{t_\alpha:\alpha\in Z\}$ is an antichain, as if $\alpha\neq\beta$ are 
in $Z$, then some $p'\leq p_\alpha,p_\beta$ forces that $t_\alpha$, 
$t_\beta$ are both in $A$. Contradiction to the fact that $T$ is Suslin 
in the ground model. 

*In Jech's above mentioned book, there is a forcing adding a Suslin tree 
with finite conditions (in the exercises). Force with it over a model 
of non-CH.  
In K. Devlin: $\aleph_1$ trees, Annals of Math. Logic, 13(1978), 267-330, all combinations of CH, ST (there exists a Suslin tree) and KT (there exists a Kurepa tree) and their negations are considered.
