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](https://en.wikipedia.org/wiki/Cardinal_characteristic_of_the_continuum#Bounding_number_'"<code>UNIQ--postMath-00000015-QINU</code>"'_and_dominating_number_'"<code>UNIQ--postMath-00000016-QINU</code>"') and the [dominating number](https://en.wikipedia.org/wiki/Cardinal_characteristic_of_the_continuum#Bounding_number_'"<code>UNIQ--postMath-00000015-QINU</code>"'_and_dominating_number_'"<code>UNIQ--postMath-00000016-QINU</code>"') 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 ${\frak d}^{\text{bij}}={\frak d}$?



Bounding and dominating numbers for preordering $\leq^*$ on bijections $f:\omega\to\omega$
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.


