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 $y$.)



bijections and order types
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 $\omega$. 


