1
$\begingroup$

For functions $f, g:\omega\to\omega$ we write $f \leq^* g$ if $\{x\in\omega: f(x)> g(x)\}$ is finite.

Let $S_\omega$ denote the collection of bijections $\varphi:\omega\to\omega$ Similarly to the bounding number and the dominating number respectively, we define

${\frak b}^{\text{bij}} = \min\{|B|: B\subseteq S_\omega \land \forall f\in S_\omega\; \exists b\in B(b\not\leq^* f)\}$, and

${\frak d}^{\text{bij}} = \min\{|D|: D\subseteq S_\omega \land \forall f\in S_\omega\; \exists d\in D(f\leq^* d)\}$.

Do we have ${\frak b}^{\text{bij}}={\frak b}$? And what about ${\frak d}^{\text{bij}}={\frak d}$?

$\endgroup$
3
$\begingroup$

Of course $\mathfrak{b}^\mathrm{bij} \geq 2$ and it is not hard to see that $\{id_\omega,f\}$ is an unbounded family if $f$ is the function that permutes each even number with its successor. So $\mathfrak{b}^\mathrm{bij} =2$.

I suspect $\mathfrak{d}^\mathrm{bij}= \mathfrak{c}$ but I don´t have time to check this.

$\endgroup$

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.