2
$\begingroup$

Background

The fact that there is no suborder of $\mathbb R$ which is of type $\omega_1$ suggests (to me) that the continuum $c$ cannot be very far from $\omega_1$: How could $c$ be far away from $\omega_1$ if there is no room for an order embedding of $\omega_1$ in $\mathbb R$? Of course, this fact is a consequence of the separability of $\mathbb R$ (which is itself an amazing fact: How can continuum many aligned irrationals be separated by only countably many rationals?)

From the idea that the continuum cannot be very far from $\omega_1$ because there is no room in $\mathbb R$ to embed $\omega_1$, one can easily formulate an axiom implying $CH$:

Preliminary definition

Let $\kappa$ be an infinite cardinal and $L$ be a total order. We say that $L$ is $\kappa$-unbounded if $|L|=\kappa$ and for every $a\in L$, we have that $|\left\{x\in L : a < x\right\}|=\kappa$.

The axiom

If $\kappa$ and $\lambda$ are infinite cardinals, $\lambda<\kappa$, and $L$ is a $\kappa$-unbounded total order, then there is an order embedding $f: \lambda\rightarrow L$ (in other words, there is a suborder of $L$ of type $\lambda$).

I have not double-checked every detail, but I am convinced that this axiom easily implies $GCH$ for at least all strong limit cardinals. In particular, it implies $CH$.

Question

Is the above axiom consistent with $ZFC$? Maybe there is an easy counterexample, but I have not found one. I know that this is related to the dense set problem as presented in Baumgartner, J., Almost disjoint sets, the dense set problem and the partition calculus.

EDIT

In view of Goldstern's counterexample, here is a modification of the axiom that might be consistent with ZFC:

Assume that $\kappa$ and $\lambda$ are infinite cardinals, $\lambda<\kappa$, $L$ is a $\kappa$-unbounded total order and that $L^*$ (the reverse order) is also $\kappa$-unbounded. Then there is an order embedding $f: \lambda\rightarrow L$ or an order embedding $g: \lambda^*\rightarrow L$ (in other words, there is a suborder of $L$ of type $\lambda$ or of type $\lambda^*$).

This still implies $CH$ and the given counterexamples do not apply.

$\endgroup$
  • 2
    $\begingroup$ Trivial counterexample: $\omega^*$ ($\omega$ reversed) does not embed into $\omega_1.$ $\endgroup$ – Elliot Glazer Oct 21 '19 at 6:29
  • 1
    $\begingroup$ @ ElliotGlazer This is not a counterexample, however Goldstern has provided a simple one. $\endgroup$ – Rodrigo Freire Oct 21 '19 at 11:27
  • 3
    $\begingroup$ Rodrigo, @Elliot means by this that the argument "since $\omega_1$ does not embed into the reals, the two cannot be too far from each other" is entirely false. $\endgroup$ – Asaf Karagila Oct 21 '19 at 18:04
  • 2
    $\begingroup$ Yeah, the motivation is shaky. But something nice about the new formulation is that it's a weakening of one characterization of weakly compact cardinals (uncountable $\kappa$ is weakly compact iff $\kappa^*$ or $\kappa$ embeds into every ordering on $\kappa$). $\endgroup$ – Elliot Glazer Oct 21 '19 at 18:07
  • 2
    $\begingroup$ Note that from Erdös-Rado one can conclude the following: If $\kappa$ is a strong limit, then for every $\lambda<\kappa$, every linear order of cardinality $\kappa$ will contain a copy of $\lambda$ or $\lambda^*$. $\endgroup$ – Goldstern Oct 21 '19 at 18:42
6
$\begingroup$

Your axiom is inconsistent. (Or perhaps I have misunderstood it.)

Let $L_n:= \aleph_n$ with the reverse order, and let $L:= L_1 + L_2 + \cdots$ (horizontal sum); equivalently, let $L$ be the lexicographic order on $\bigcup_k \{k\}\times L_k$. Then $L$ is $\aleph_\omega$-unbounded, yet there is no order-preserving embedding of $\omega_1$ into $L$: (EDITED to simplify:) Every well-ordered subset of $L$ is finite in each $L_n$, hence at most countable.

| cite | improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ Very nice! It is quite simple, if we just take each $L_n$ in your construction to be $\omega_2$ reversed, then it will also work. $L$ is $\omega\times\omega_2^*$, it is $\omega_2$-unbounded in this case and one cannot embed $\omega_1$ in $L$. Thanks. $\endgroup$ – Rodrigo Freire Oct 21 '19 at 11:26
  • 2
    $\begingroup$ However, the axiom could be modified to exclude these counterexamples, still implying $CH$. It is enough to require $L$ reversed to be also $\kappa$-unbounded and to ask just for an embedding of $\lambda$ or of $\lambda^*$. $\endgroup$ – Rodrigo Freire Oct 21 '19 at 12:53
4
$\begingroup$

Partial answer for the newly formulated axiom, which I'll just call Ax.

First, Ax does imply GCH. Assume Ax and suppose GCH first fails at $\kappa.$ Let $L=(2^{\kappa} \setminus \{\alpha \mapsto 0, \alpha \mapsto 1\},<),$ where $<$ is the lexicographic ordering. Then $\kappa^+$ embeds into $L$ by Ax and the symmetry of the order. Notice the set $S$ of eventually 0 functions is dense in $L,$ and $|S|=\kappa$ by GCH below $\kappa.$ But there is an injection from $\kappa^+$ into $S,$ contradiction.

Now GCH implies many cases of Ax, including $\kappa$ successor, weakly compact, or countable cofinality (clarification: I'm using $\kappa$ to refer to the smaller cardinal rather than $\lambda,$ which is not how the variables are defined in the question). In fact, in each of these cases, $\kappa$ has the stronger property that $\kappa$ or $\kappa^*$ embeds into each ordering on $\kappa^+.$

First, for $\kappa=\omega$ or weakly compact $\kappa,$ it is known they have the stronger property that $\kappa$ or $\kappa^*$ embeds into each ordering on $\kappa.$

For $\kappa$ a successor cardinal, the result is immediate from Erdős-Rado and GCH.

For $\kappa$ singular of countable cofinality, fix an ordering $L=(\kappa^+, <').$ Fix a $\kappa$-sequence of disjoint intervals $\langle (\alpha_{\xi}, \beta_{\xi}), \xi<\kappa \rangle$ such that $|(\alpha_{\xi}, \beta_{\xi})| \ge \kappa$ for all $\xi.$ (*) For each $\xi,$ either $\lambda$ embeds into $(\alpha_{\xi}, \beta_{\xi})$ for each $\lambda<\kappa$ or $\lambda^*$ embeds into $(\alpha_{\xi}, \beta_{\xi})$ for each $\lambda<\kappa$ (otherwise we get a counterexample to the successor case). By restricting to a $\kappa$-sequence of $\xi_{\eta}$ and flipping $<'$ if necessary, we may assume wlog that each $\lambda<\kappa$ embeds into each $(\alpha_{\xi}, \beta_{\xi}).$

Applying Erdős–Dushnik–Miller to the projection of $<'$ onto the intervals, there is an ascending $\omega$-sequence of intervals or descending $\kappa$-sequence. In the first case, since $\text{cf}(\kappa)=\omega,$ this gives us an embedding of $\kappa$ into $L.$ In the second case, there is clearly an embedding of $\kappa^*$ into $L.$

(*) I'm not sure this sequence exists, see comments discussion.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Could you say a bit more about the successor case? Prima facia, Erdos-Rado only says something about colorings of pairs on double successors under GCH. If we use only two colors then the argument can be carried out for successors of regulars generally, but I am not sure about successors of singulars. $\endgroup$ – Monroe Eskew Oct 22 '19 at 4:16
  • $\begingroup$ For example, fix a scale $\langle f_\alpha : \alpha < \aleph_{\omega+1} \rangle$. For $\alpha < \beta$, let $c(\alpha,\beta)$ be the least $n < \omega$ such that $f_\alpha(n) < f_\beta(n)$. There cannot exist a homogeneous set of size $\aleph_\omega$. $\endgroup$ – Monroe Eskew Oct 22 '19 at 4:24
  • $\begingroup$ @MonroeEskew I think I'm using variables inconsistently with the original question. In my casework, $\kappa$ is representing the smaller cardinal (what $\lambda$ is in the problem statement). I'll clarify in my answer. $\endgroup$ – Elliot Glazer Oct 22 '19 at 4:26
  • $\begingroup$ In your argument for $\kappa$ singular, why do we have a $\kappa$-sequence of disjoint intervals? $\endgroup$ – Monroe Eskew Oct 22 '19 at 4:39
  • $\begingroup$ For any interval with $\kappa^+$ elements, there is a cut such that one side has $\kappa^+$ elements and the other has at least $\kappa$ elements. Recursively cut out an interval with $\kappa$ elements while leaving $\kappa^+$ elements. At the $\alpha$th stage, for limit $\alpha<\kappa,$ the $(\alpha_{\xi}, \beta_{\xi})$ constructed thus far partition the remaining $\kappa^+$ elements into less than $\kappa$ many intervals, one of which has $\kappa^+$ elements, and use that interval to continue the construction. $\endgroup$ – Elliot Glazer Oct 22 '19 at 4:49
2
$\begingroup$

It seems to me that the axiom is equivalent to GCH:

Assume GCH. Let $\lambda<\kappa$ be infinite cardinals. Let $L$ be a total order of cardinality $\kappa$. We want to prove that $L$ contains a copy of $\lambda$ or of $\lambda$ reversed.

Case 1) $\kappa$ is limit. From GCH, $\kappa$ is strong limit and the result follows from Erdos-Rado.

Case 2) $\kappa$ is a successor of a successor: $\kappa=\lambda^{++}$. Then, from GCH and Erdos-Rado, $L$ contains a copy of $\lambda^+$ or of $\lambda^+$ reversed, and this is enough.

Case 3) $\kappa$ is a successor of a limit $\lambda$. From GCH and the version of Erdos-Rado given in Levy, Basic set theory, theorem 3.13, chapter IX, $\lambda^+\rightarrow (\lambda)^2_2$. Since $\kappa=\lambda^+$, we have that $L$ contains a copy of $\lambda$ or of $\lambda$ reversed, and this is enough.

EDIT

The argument can be unified: the proof of case 3) is enough.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ It seems $\kappa$-unboundedness was never needed, which gives a cleaner equivalent statement, which in turn is basically "every cardinal satisfies a weak version of weak compactness." Pretty nice. $\endgroup$ – Elliot Glazer Oct 26 '19 at 23:47
  • 2
    $\begingroup$ Yes, it was never needed, but it is also equivalent. I am calling the cardinals satisfying the weak version of weak compactness the order-replete cardinals. The nice thing to me is that the axiom is purely order-theoretic, hence it provides some geometric intuition to GCH. $\endgroup$ – Rodrigo Freire Oct 27 '19 at 0:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.