Suppose $\kappa$ is an infinite cardinal and $\alpha$ is an ordinal of cardinality $\kappa$. Is it possible to find a bijection $f : \kappa \to \alpha$ such that for all $x \subseteq\kappa$, $\mathrm{ot}(x) \leq \mathrm{ot}(f[x])$? (Here, $\mathrm{ot}(y)$ is the order type of a set of ordinals $y$.)
$\begingroup$
$\endgroup$
asked Nov 8, 2018 at 1:24
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
When $\kappa = \aleph_0$, any bijection works.
When $\kappa$ is uncountable with uncountable cofinality, there is no order-type preserving bijection $f\colon \kappa\to \kappa+\omega$. Indeed, let $X = f^{-1}((\kappa+\omega)\setminus \kappa)$. Then $X$ is not cofinal in $\kappa$, so we can pick some $\alpha\in \kappa$ greater than every element of $X$, and $f(\alpha)<\kappa$. Then $\text{ot}(X\cup \{\alpha\})>\omega$, but $\text{ot}(f(X\cup \{\alpha\})) = \omega$.
I'm not sure about the case when $\kappa$ is uncountable with cofinality $\omega$.
answered Nov 8, 2018 at 2:11
-
6$\begingroup$ I think you can proceed similarly when $\kappa$ has countable cofinality, using $\kappa+\omega_1$ instead of $\kappa+\omega$. The preimage $X$ of the final $\omega_1$ must have order-type $\leq\omega_1$ by the condition in the problem, but it can't be $<\omega_1$ by cardinality. So $X$ has order-type exactly $\omega_1$, and then $X$ can't be cofinal in $\kappa$. Then pick a larger element $\alpha<\kappa$ and notice that $X\cup\{\alpha\}$ has order-type $\omega_1+1$ while its image has order-type only $\omega_1$, just as in your proof. $\endgroup$ Commented Nov 8, 2018 at 2:58
-
$\begingroup$ @AndreasBlass Ah great, thanks for supplying the argument. $\endgroup$ Commented Nov 8, 2018 at 3:00