We say $S$ is a Suslin forest if adding a minimum to $S$ we have a Suslin tree. So a Suslin Forest is essentially a Suslin tree $S$ in which we drop the requirement for $S$ to have a single root. Notice that every uncountable subset of a Suslin tree is a Suslin forest.

Given a Suslin forest $S$, I would like to know wether it is possible or impossible to define a function $f:S\to S$ such that for every $x,y\in S$ with $x<y$ we have $f(x)\perp f(y)$ (i.e. $f(x)\nleq f(y) \land f(x)\ngeq f(y))$ and a proof or reference to a proof.

We say $S$ is a non-special Aronszajn forest if adding a minimum to $S$ we have a non-special Aronszajn tree.

Would the same result holds for a $S$ non-special Aronszajn forest which is not necessarily a Suslin forest?

EDIT: For S Suslin forest, $f(S)$ is still a Suslin forest. For $S$ non-special Aronszajn forest which is not necessarily Suslin forest, we would like to require also that $f(S)$ is still a non-special Aronszajn.

  • $\begingroup$ Usually having a single root is part of the definition of "tree", which itself is part of the definition of a Suslin tree. Of course, it's not a significant one, we can always add it "by force". Yes, it makes things a bit more awkward when you do these things, because then the root goes somewhere and it must be compatible with something. But that's why we forego the root in the requirement. Just like how defining a regressive function will generally require it to be regressive for non-zero ordinals (or: where possible). $\endgroup$
    – Asaf Karagila
    Jul 12, 2018 at 10:25
  • $\begingroup$ Yes exactly. For the problem that made me think about this, S was an uncountable subset of a Suslin tree. This should be a "Suslin tree" with not necessarily one root. If this can make the question clearer, I can edit the question to state that S is just an uncountable subset of a Suslin tree. $\endgroup$
    – Cla
    Jul 12, 2018 at 10:39
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    $\begingroup$ If $S$ is a non-special Aronszajn trees which is not Souslin, then one may take an antichain $A$ in $S$ of size $\omega_1$ and just define a bijection $f: S \to A.$ So the main interest is when $S$ is Souslin. $\endgroup$ Jul 12, 2018 at 10:56
  • $\begingroup$ Thanks for the remarks. I edited the question, I hope now is clearer. $\endgroup$
    – Cla
    Jul 12, 2018 at 11:17

1 Answer 1


Yes, this can happen, and indeed it happens in a subforest of any given Souslin tree.

Let's start with an illustrative case. Sometimes people consider Suslin trees that are not necessarily normal, and where for example, a sequence converging to a limit level can have more than one bounding node at that level.

Let us consider such a Suslin tree $T$, where every node is a binary splitting node, and furthermore, every limit node has a partner limit node with the same predecessors. Such a tree would arise, for example, from a normal $2$-splitting Suslin tree, simply by omitting the nodes at limit levels, in effect making the two successors of that limit node the new successors of the branch below.

Now let $S$ consist of two copies of $T$. This is a Suslin forest. Notice that every element of $S$ can be determined by a binary sequence of some successor ordinal length $\alpha+1$. Namely, the first bit tells you which copy you are in; the middle bits tell you about branching; and the limit bits tell you which of the two limit nodes you're at.

Let $f:S\to S$ be the function arising by simply swapping the last bit of the binary representing sequences.

It follows that if $x<y$ in $S$, then $f(x)\perp f(y)$, since only the last bit is changed. In other words, $f(y)$ extends $x$, which is incomparable with $f(x)$.

The idea works generally, if one is willing to pass to a subforest.

Theorem. Every Suslin tree $T$ has a Suslin subforest $S\subset T$, such that there is a function $f:S\to S$ with $x<y\to f(x)\perp f(y)$.

Proof. I claim that there is a Suslin subforest $S\subset T$ such that every node of $S$ has another node on the same level, with the same predecessors. One can pick $S$ inductively: pick two incomparable nodes to be the roots; for every node you've picked, pick two incomparable successor nodes; and at limits, pick two incomparable nodes to serve as upper bounds. (One could also pick more than two, if desired.) Alternatively one can prune $T$ above the two roots by removing the limit level or non-branching nodes. This defines the forest $S$, which is Suslin because any chain or antichain in $S$ would give rise to a chain or antichain in $T$.

The key property of $S$ is that every node $x$ in $S$ has another node $f(x)$ on the same level as $x$ and with the same predecessors as $x$. This property is sufficient to know that $x<y$ implies $f(x)\perp f(y)$, because $x$ will be below $f(y)$, but at the same level as $f(x)$ and incomparable with $f(x)$. $\Box$

One can consider the same operation on the complete binary tree $2^{<\omega}-\{\langle\rangle\}$, without the empty sequence. If you flip the last bit of every sequence to its opposite, then you have $x<y\to f(x)\perp f(y)$. The function $f$ above is in effect simply doing this to each condition in the finite part above the largest limit level preceeding a given node.

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    $\begingroup$ I see a downvote on this (to my view good) answer which I don't understand properly. I will appreciate if one adds some explanatory remarks clarifying the reason behind it. $\endgroup$ Jul 13, 2018 at 14:20
  • $\begingroup$ Perhaps someone objects that the construction does not work with normal Suslin trees, which have unique limit nodes. That would seem to be a quite reasonable objection, and I would be interested to know whether one can find such a function generally for Suslin trees with the normality condition. $\endgroup$ Jul 13, 2018 at 14:49
  • $\begingroup$ Thank you this is a really clear answer. For my purpose what you showed should be enough, but yes it would be interesting to see also a result for Suslin trees in the general case, so I might wait a bit to close the question. $\endgroup$
    – Cla
    Jul 14, 2018 at 9:06
  • $\begingroup$ If given S Sulin we can find a splitting non-normal tree T as the one you described such that S is isomorphic to a subset of T and T is isomorphic to a subset of S then we're done, since a function from T to T can then be restricted on the domain to S, and then the image can be projected inside S. I can't figure out if this is useful or there are no trees for which this could be done. $\endgroup$
    – Cla
    Jul 14, 2018 at 11:35

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