1

Recently in chat, we investigated the explicit construction of $\omega_1$. Regardless on whether we use ZFC or hartogs number, we seemed to hit a roadblock.

Using either ZFC or hartogs number, we managed to came up a set of countable ordinals $S$ that corresponds to well orderings on $\omega$. However, other than its existence, there seemed to be no algorithmic way to describe that well ordering in more detail

(for example, some explicit but not necessary computable predicate P(x,y) such that given any pair (x,y) it unambiguously give either true or false to the question x < y, that is, $P(x,y)$ is not of the form "There exists a well ordering...")

therefore cannot show that $S$ is well ordered and hence $\omega_1$ cannot be shown to exist by construction.

  1. Is there exists a predicative well ordering of an uncountable set $S$ which the injection $\omega \to S$ can be constructed without first assuming $S$ can be well ordered without the axiom of choice (and possibly without excluded middle)?
  2. If there is, what is $P$, that is, the criteria on how exactly any given countable well orderings are well ordered within $S$?
Secret
  • 2,420
  • There is an explicit construction of $\omega_1$ as an element of $P(P(\Bbb Q))$ (where $P$ denotes the power set), using just ZF and $\Bbb Q$. Would that be the kind of thing you're looking for? – Arthur Oct 06 '17 at 16:12
  • But how can we be sure that $P(P(\Bbb{Q}))$ will be well ordered since AC is missing. And I vaguely remember that axiom of powerset does not guarentee well ordering (e.g. construct $\Bbb{R}$ from $\Bbb{Q}$)? – Secret Oct 06 '17 at 16:15
  • 1
    We can't. But some elements of it can be well-ordered, and at least one of those is order isomorphic to $\omega_1$. – Arthur Oct 06 '17 at 16:16
  • @Arthur: Why rationals, and not just naturals? The standard construction is to consider the equivalence classes of countable well-orderings. ZF can prove that these are well-ordered under embedding. But that proof relies on LEM on "There is an embedding of $x$ into $y$." given any two countable well-orderings $x,y$. – user21820 Oct 06 '17 at 16:17
  • But $\omega_1$ is defined to be the well ordering of all countable well orderings, if this set is only partially well ordered, then we cannot say this set is order isomorphic to $\omega_1$ (also $\omega_1$ will not contain itself as per definition of ordinals, thus if we can found an element in it that is isomorphic to $\omega_1$, then $\omega_1$ will be countable and result in contradicting that $\omega_1$ contains all countable well orderings)? – Secret Oct 06 '17 at 16:18
  • 1
    @user21820 I use the rationals because the construction is more human readable and uses the standard total ordering on the rationals for all its worth. One could use any countably infinite set, but it wouldn't be as transparent. – Arthur Oct 06 '17 at 16:19
  • 1
    Oh, and I made a mistake. There is a subset of $P(P(\Bbb Q))$ which is well-orderable and order isomorphic to $\omega_1$. – Arthur Oct 06 '17 at 16:22
  • @Arthur In $\mathsf{ZF}$, there is a surjection from $\mathcal P(\mathbb N)$ onto $\omega_1$, so there is an injection of $\mathcal P(\omega_1)$ into $\mathcal P(\mathcal P(\mathbb N))$. – Andrés E. Caicedo Oct 06 '17 at 17:57
  • "the inability to write down computable ordinals $x,y$ between $\omega_1^{CK}$ and the large Veblen ordinal means that given arbitrary $x,y$ we cannot justify that an injection exists from one into another." I don't understand this - can you explain more clearly what the problem is here? Given two ordinals $\alpha,\beta$ we define an injection from one to the other via transfinite recursion along them - and I believe this is predicative (the crucial point being that we're given $\alpha$ and $\beta$ in this situation, we don't have to build them predicatively). – Noah Schweber Oct 06 '17 at 19:03
  • How is $\alpha$,$\beta$ given? If $\beta$ is a successor of $\alpha$, then I can see we can show that $\alpha < \beta$ by injection. But if $\beta$ is a limit or an ordinal that is not a successor of $\alpha$, how can we show that $\alpha <\beta$ without having to assume that our given $\alpha$ injects into $\beta$? Or is the Hartogs number construction guarantee that for all $\alpha,\beta$, $\alpha < \beta$ and thus established that $S$ is a well ordering even though we cannot construct $\alpha,\beta$? – Secret Oct 07 '17 at 04:52
  • @NoahSchweber: The chat discussion Secret was talking about was actually one with me, where I claimed that the LEM I stated above was a crucial ingredient of the proof that given any two computable ordinals one embeds into the other. I don't see much philosophical problem if we assume that LEM instance. The problem with it is that it assumes that "embedding" is somehow a fixed concept. In ZFC it indeed is fixed by the power-set axiom; $\mathcal{P}(α \times β)$ is effectively an indicator function on the entire universe. But the power-set is not predicative. – user21820 Oct 07 '17 at 10:48
  • @Secret: The large Veblen ordinal $v$ is actually a red herring here, because if you believe it is well-founded then you can simply use its syntactic representation as the ordinal notation for which you can construct a computable ordering of everything below it, and hence can extend it upward easily. For example, you could use pairs of ordinals less than $v$, which gets you a computable ordering of length $v·v$. – user21820 Oct 07 '17 at 10:53
  • If $1=0$ is not a theorem of ZFC then we cannot prove the existence of $\omega_1$ from ZFC$-$P (where P is the Power-Set Axiom) because in ZFC it is demonstrable that the set $H_{\omega_1}$ of hereditarily countable sets is a model for (ZFC$-$P+($\neg$ P)+($\forall x; (|x|\leq \omega$))). – DanielWainfleet Oct 24 '17 at 12:11
  • 1
    If you can prove that any predicative function applied to a countable ordinal generates a countable ordinal, then the answer is obviously not. Since $\omega_1$ is regular (at least assuming some bits of choice), any countable-to-countable function on ordinals would require $\omega_1$ iterations to reach $\omega_1$, which is exactly what being impredicative is all about. – Asaf Karagila Oct 25 '17 at 09:44
  • Not sure if I mistaken again, yes $\omega_1$ and $\aleph_1$ are regular when at least countable choice is assumed. However I recall in choiceless ZF, there exists models where $\aleph_1$ is singular (I am not sure yet if we can explicitly pick those, however), because countable unions of countable sets can be uncountable, but I am not sure if that will also make $\omega_1$ singular, thus allowing countable iterations of countable to countable functions to reach $\omega_1$. – Secret Oct 25 '17 at 10:34
  • 1
    $\aleph_1$ and $\omega_1$ are the same object. The former is used to hint "we are talking about cardinals and cardinal arithmetic" and the latter denotes "we are talking about ordinals and order types". So $\aleph_1+\aleph_0=\aleph_1$, but $\omega_1+\omega\neq\omega_1$. Now. It is indeed consistent with ZF that $\omega_1$ is singular and thus the countable union of countable sets. But again, as far as predicativity, you would need to somehow argue that you can find a countable cofinal sequence which is obtained by iterating some predicative construction in a predicative way. [...] – Asaf Karagila Oct 25 '17 at 11:10
  • 1
    [...] If you work under some reasonable assumptions that predicative definitions are absolute between transitive models of ZF, then this puts a big stick into the wheels of your idea, since $\omega_1$ while singular in your universe of ZF is not the $\omega_1$ that you find in L, an inner model of choice, thus the predicative construction cannot even surpass a very fairly specific countable ordinal as far as your universe is concerned. So the whole approach collapses. – Asaf Karagila Oct 25 '17 at 11:12
  • I see, I think I understood the problem better now, thanks guys! – Secret Oct 25 '17 at 11:47

0 Answers0