Is not SH + not CH consistent?

Question

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?

Answer 1

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...)

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

2. Saharon Shelah, Can you take Solovay’s inaccessible away?, Israel J. Math. 48 (1984), no. 1, 1--47.

Answer 2

Two more solutions.

1. 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.

2. 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.