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Everyone knows that the real line $\langle\mathbb{R},<\rangle$ is the unique endless complete dense linear order with a countable dense set. Suslin's hypothesis is the question whether we can replace separability in this characterization with the assertion that the order has the countable chain condition, that is, that every set of disjoint intervals is countable. In other words, Suslin's hypothesis, in the original formulation, is the assertion that the real line is the unique endless complete dense linear order with the countable chain condition.

Question. Does ZF plus the axiom of determinacy AD imply the original Suslin hypothesis?

Set theorists proved in ZFC that Suslin's hypothesis is equivalent to the assertion that there is no Suslin tree, which is a tree of height $\omega_1$ with no uncountable chains or antichains. And ZF plus the axiom of determinacy AD settles this version of SH by proving that $\omega_1$ is measurable and hence weakly compact and hence has the tree property and so under AD there is there is no Suslin tree.

So under AD there is no $\omega_1$-Suslin tree. But does this mean that there is no Suslin line?

The point is that refuting a Suslin tree does not seem directly to refute all Suslin lines in the non-AC context, and therefore it does not seem to settle the original Suslin problem under AD. So the question is: does ZF+AD settle the original Suslin problem?

Please feel free to post an answer explaining the precise details of the argument that AD implies there is no $\omega_1$-Suslin tree.

I heard this question this evening at a party at a conference in honor of Simon Thomas from a certain prominent set theorist, aged Scotch in hand, who told me that he would rather not be mentioned, but who said he was fine for the question to be posted.

Let me update the question (after Asaf's very nice answer) to ask about the situation where we also have the axiom of dependent choice DC. And perhaps one really wants to know about the case in $L(\mathbb{R})$ under AD.

Question. Does ZF+AD+DC imply the original Suslin hypothesis?

Question. Does $L(\mathbb{R})$ have a Suslin line assuming AD?

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    $\begingroup$ @AsafKaragila Does the argument need a normal measure? The one I have in mind just replicates the proof of König's lemma at $\omega$, building a branch through the tree by finding on each level a node with measure one many descendants in the tree. I think completeness is enough for this although, looking at it now, it seems to use dependent choice, so perhaps the argument doesn't work regardless. $\endgroup$ – Miha Habič Sep 17 '17 at 7:21
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    $\begingroup$ A cautionary note: There are two ways to formulate the tree property for $\omega_1$. One talks about trees with $\omega_1$ nodes; the other about trees with $\omega_1$ levels. Since each level is countable, these are equivalent in ZFC, but I don't see the equivalence in just ZF. The proof that measurability implies the tree property is fine under the first formulation, but I don't see how to do it under the second. $\endgroup$ – Andreas Blass Sep 17 '17 at 13:31
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    $\begingroup$ Yes, my comment about measurable$\to$tree property$\to$no Suslin trees was too quick, since there are several subtle issues there. Let me reiterate my request for someone to post an answer showing that AD refutes Suslin trees. $\endgroup$ – Joel David Hamkins Sep 17 '17 at 14:50
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    $\begingroup$ Hi Joel, you may already be aware of this, but I saw on the arXiv this morning that your last question has been answered: arxiv.org/pdf/1803.08201.pdf. $\endgroup$ – Will Brian Mar 23 '18 at 13:02
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    $\begingroup$ @JoelDavidHamkins The question is now answered in L(R ) with Determinacy Satisfies the Suslin Hypothesis $\endgroup$ – Mohammad Golshani Mar 24 '18 at 8:05
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No. Because of silly reasons.

Recall that the powers of $\sf AD$ are quite limited to the world below $\Theta$. In particular, the proof that $\sf AD$ does not imply countable choice goes through adding a Dedekind-finite set somewhere very high up the von Neumann hierarchy. Where the reals cannot interfere.

Simply replicate this proof, but this time embed a copy of the Mostowski linear ordering, instead of any ol' Dedekind-finite set. This would result in a linear ordering whose order topology is strongly connected (and hence the order is complete), where every subset is a finite union of intervals, and the power set is Dedekind-finite. So in particular, every collection of pairwise disjoint intervals is finite.

(Note that to get Mostowski's ordering you'd need each point to be a set of sets of ordinals, rather than just a set of ordinals, so you need to force with something like $\operatorname{Add}(\kappa,\kappa\times\omega)$, and then group your generic subsets into collections of generic subsets indexed by the rational numbers, and take permutations which turn each "rational" into a sufficiently-non well-orderable set, and permutes the "rationals" in an order preserving way.

Of course, using any other method of adding generic sets, not necessarily Cohen subsets, also works for these purposes.)


Here is an outline of how to construct such model.

Start with a model of $\sf ZFC$ with $\omega$ Woodin cardinals, and for good posterity also $\sf GCH$. Pick some regular $\kappa$ much much larger than all the Woodin cardinals. For example, the double successor of their limit.

Next step, take the product of the two symmetric systems: (1) add a Mostowski order above $\kappa$, and let $\cal A$ denote the Mostowski ordering, along with its linear order; (2) collapse the Woodin cardinals to force $\sf AD$ in an inner model.

Finally, in the symmetric extension consider $L(\Bbb R,\cal A)$. By the closure of adding Cohen subsets to $\kappa$, and it being large enough, we have ensured that the set of reals in this models are all definable from reals, i.e. live in $L(\Bbb R)$. Therefore $\sf AD$ holds. At the same time, all the relevant arguments about the Mostowski ordering involve its power set being Dedekind-finite, which is true in the symmetric extension, and therefore in any inner model which knows the underlying set.


The answer is still negative even if one requires $\sf DC$ to hold. Indeed, we can generalize the above argument in the following way (which now requires assuming $\sf GCH$, or something similar about the continuum at regular cardinals):

  1. By $\sf GCH$, $\eta_\kappa$ exists, which is a $\kappa$-saturated dense linear ordering of size $\kappa$. Therefore it is also homogeneous.

  2. Instead of adding $\kappa\times\omega$ subsets of $\kappa$, we add $\kappa\times\kappa$, where the second $\kappa$ is indexed using a linear ordering isomorphic to $\eta_\kappa$. Now we take the filter of subgroups to be generated by fixing countably many points, rather than finitely many.

  3. The technique and argument now give us that if $A$ is the generic linear ordering, every subset of $A$ is the countable union of intervals. This, along with the saturation of $\eta_\kappa$, readily implies that the order is complete.

    Given any cut, we can write it as the union of two intervals, they have a common support which is a countable set of points on the line. One can show that these points must have a cofinal sequence in the upper and lower parts of the cut (otherwise there is a "gap" between the cuts which can be moved by a suitable automorphism). And by saturation we can realize the type of the cut, and thus it is an endpoint of one of the parts of the cut.

  4. The linear order we added is actually ccc, since any uncountable family of pairwise disjoint intervals would have a union which cannot be decomposed into only countably many disjoint intervals. Of course, the order is not separable, since $\kappa$ is just too darn big, and separability would imply that we collapsed $\kappa$ to be the continuum in the outset $\sf ZFC$ model, but $\kappa$ is in fact the successor of the continuum (or larger!), so that cannot be.

  5. In $L(\Bbb R,\cal A^\omega)$, where $\cal A$ is the ordered set, as computed inside the symmetric extension of both the collapses of the Woodin cardinals, and adding the Mostowski $\sigma$-order, we get that $\sf ZF+DC+AD$ holds. But $\cal A$ is still a non-separable ccc and complete dense linear ordering. Therefore a counterexample.


So it seems like the real question is assuming that $L(\Bbb R)\models\sf AD$; does it also satisfies $\sf SH$? Or, perhaps, does $\sf ZF+AD+\mathit{V=L(\Bbb R)}\implies SH$?

Here I'm a bit more skeptical about how much the above ideas could work. And it actually seems to be plausible that there are no Suslin trees, in the broad sense of the word.

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  • $\begingroup$ I'd expect, however, $\sf AD$ to solve the Suslin Hypothesis for sets below $\Theta$, or you know, below the power set of the reals. $\endgroup$ – Asaf Karagila Sep 17 '17 at 7:05
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    $\begingroup$ Parson my ignorance, but what is $\theta$? $\endgroup$ – David Roberts Sep 17 '17 at 8:52
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    $\begingroup$ The least ordinal such that there is no surjection from the reals onto it. What I would normally call the Lindenbaum number of the reals. $\endgroup$ – Asaf Karagila Sep 17 '17 at 8:57
  • $\begingroup$ Thanks very much for this post! This seems to answer the question I had asked. But in truth, I had had in mind the case where we would also have DC, which I didn't state, and it seems that your technique would not work in that case. $\endgroup$ – Joel David Hamkins Sep 17 '17 at 11:33
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    $\begingroup$ @NotMike: Yes, you're right of course, I thought about the switch from a tree to a line wrong. Actually looking back, I'm not entirely sure what I meant by below $\Theta$. Maybe in $V_\Theta$ or something. $\endgroup$ – Asaf Karagila Sep 19 '17 at 17:12
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Nice question! But which aged set theorist drinks prominent Scotch?

The following does not answer this question either. Let $T \in L$ be your favorite Suslin tree of $L$. Consider the $L({\mathbb R})$ of $L[g]$, where $g$ is generic for adding $\omega_1$ Cohen reals. $T$ is still a Suslin tree in $L[g]$. The set of all maximal branches thru $T$ is then a Suslin line in $L({\mathbb R})^{L[g]}$, and $L({\mathbb R})^{L[g]}$ is a model of ZF plus DC plus "there is no w.o. of the reals." So a Suslin line doesn't give a w.o. of ${\mathbb R}$.

As has been poited out, this approach can't be generalized to show that in the presence of large cardinals (or just ${\sf AD}^{L({\mathbb R})}$), $L({\mathbb R})$ has a Suslin line, as then no $T \in M$, $M$ any inner model of $L({\mathbb R})$ satisfying choice, can be an Aronzsajn tree in $L({\mathbb R})$.

But, assuming large cardinals, the question if $L({\mathbb R})$ has a Suslin line should have an answer as ${\sf ZFC}$ is $\Omega$-complete concerning truth in $L({\mathbb R})$.

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Concerning the existence of Suslin trees under $\mathsf{AD + V = L(\mathbb{R})}$:

Assume $ZF + AD^+ + V = L(\mathscr{P}(\mathbb{R}))$. Let $(T,\prec)$ be an $\omega_1$-tree with all levels countable.

(I) Suppose $T$ is wellorderable. We can repeat Hamkin's argument above: One can show $T$ is in bijection with $\omega_1$. $(T,\prec)$ can be coded as a subset of $\omega_1$. $(T,\prec)$ is constructible from real $r$ by a result of Solovay; hence, $(T,\prec) \in L[r]$. $r^\sharp$ exists, so $\omega_1^V$ is weakly compact and has the tree property. Therefore, in $L[r]$, $T$ has a branch which is in bijection with $\omega_1^V$. $(T,\prec)$ is not a Suslin tree in $V$.

(II) Suppose $T$ can not be wellordered. By Theorem 1.4 of "A Trichotomy Theorem in Natural Models of $\mathsf{AD}^+$" by Caicedo and Ketchersid, there is an injection $\Phi : \mathbb{R} \rightarrow T$ under these assumptions. Define a prewellordering on $\mathbb{R}$ by $x \sqsubseteq y$ if and only if the level of $(T,\prec)$ that $\Phi(x)$ belongs to is less than or equal to the level of $(T,\prec)$ that $\Phi(y)$ belongs to. Since $(T,\prec)$ has countable levels, $\sqsubseteq$ is a prewellordering on $\mathbb{R}$ with each prewellordering class countable. This does not exists under $\mathsf{AD}$.

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