A $P$-point is an ultrafilter $\scr U$on $\omega$ such that for every function $f:\omega\to\omega$ there is $x\in {\scr U}$ such that the restriction $f|_x$ is either constant, or finite-to-one.

A $Q$-point is an ultrafilter $\scr U$on $\omega$ such that for every function $f:\omega\to\omega$ with the property that $f^{-1}(\{m\})$ is finite for each $m\in \omega$, there is $x\in {\scr U}$ such that the restriction $f|_x$ is injective.

$P$-points need not exist, and $Q$-points need not exist.

Question. Is it possible that neither $P$- nor $Q$-points exist?


Your definition of P-point should end with "finite-to-one" rather than "injective". As it stands, it defines selective ultrafilters, not P-points.

As far as I know, the consistency of the statement "neither P-points nor Q-points exist" is still an open problem. It is known that this statement implies $2^{\aleph_0}\geq\aleph_3$.

EDIT in response to a request for a reference for the last statement: Although the statement is explicitly somewhere in the literature, I can't find it just now, so instead I'll indicate how it follows from two results that I can find, Theorems 9.25 and 9.27 in my chapter of the Handbook of Set Theory. Neither result is due to me; references are given in the chapter, a pre-publication version of which is available at http://www.math.lsa.umich.edu/~ablass/hbk.pdf .

The results involve the cardinal characteristics cov$(\mathcal B)$ [the smallest number of meager sets to cover $\mathbb R$], $\mathfrak d$ [the smallest number of functions $\omega\to\omega$ to dominate all such functions], and $\mathfrak c$ [the cardinal of the continuum]. It is provable in ZFC that $$ \aleph_1\leq\text{cov}(\mathcal B)\leq\mathfrak d\leq\mathfrak c. $$ Theorem 9.25 says (among other things) that if $\mathfrak d=\mathfrak c$ then there exists a P-point. Theorem 9.27 says (among other things) that if cov$(\mathcal B)=\mathfrak d$ then there exists a Q-point. So in order to have neither a P-point nor a Q-point, you'd need $$ \aleph_1\leq\text{cov}(\mathcal B)<\mathfrak d<\mathfrak c, $$ which implies $\mathfrak c\geq\aleph_3$.

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    $\begingroup$ Do you have a reference for the last claim? $\endgroup$ – Wojowu Mar 24 at 21:17
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    $\begingroup$ @Wojowu I've edited the answer to include a proof from results with references. $\endgroup$ – Andreas Blass Mar 25 at 0:58

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