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I had previously posted an answer, which was not correct. The correct answer is negative, as shown in the first part of the answer of Calliope Ryan-Smith, following a suggestion of Andreas Leitz there …

The answer is yes, because every forcing notion is equivalent to a forcing notion with finite predecessors. Theorem. … Every forcing notion is forcing equivalent to a forcing notion with finite conditions, a family of finite sets ordered by (reverse) inclusion. …

The paper shows how to have a supercompact cardinal $\kappa$, say, that is indestructible by certain kinds of forcing but not others. …

In ZFCA, every set is equinumerous with a pure set, since every well ordering of a set makes it bijective with an ordinal, which is pure. Therefore the cardinal structure of any model of ZFCA is refle …

Let $P$ arise from product forcing $Q\times Q$. So forcing with $P$ adds two mutually generic filters for $Q$, one on each factor. …

This forcing is equivalent to adding a Cohen subset of $\omega_1$, which is countably closed and hence proper. … The combination of adding a Cohen set and then killing it is isomorphic to Cohen forcing, which is proper. …

Here is an example with Cohen forcing. Let $\mathbb{Q}$ be the forcing that adds two Cohen reals, viewed as binary sequences, along with their bitwise sum mod 2. … This forcing will add a Cohen real on each coordinate. …

Enumerate the dense sets $D_0$, $D_1$, and so on for Cohen forcing in $M$, forcing with finite binary sequences under end-extension. Let $c_0$ be the shortest condition in $D_0$. … Individually, they are $M$-generic Cohen reals, but they are not mutually generic for this forcing, since if we have them together we can observe the coding blocks and get $z$. …

Consider the forcing $\mathbb{P}$ that adds a generic function from $\text{Ord}$ to $V$. (One can use this forcing to force global choice — see Victoria Gitman's account.) … So we cannot so easily go to the Boolean completion of the forcing. …

In short, what I want are forcing notions $\newcommand{\P}{\mathbb{P}}\P$ and $\newcommand\Q{\mathbb{Q}}\Q$ such that every forcing extension arising from either of them is a submodel of a forcing extension … by the other, but the two forcing notions have no common forcing extension. …

Other instances are consistent, because they follow from certain forcing axioms. … forcing notions. …

This is not a full answer, but I found it interesting to notice that if we relax the c.e. requirement somewhat, then there is a sweeping positive answer. Theorem. Every complete theory extending ZFC + …

The answer to the topological version of Banach-Mazur games is negative, proved by Gabriel Debs in 1985: Debs, Gabriel, Stratégies gagnantes dans certains jeux topologiques (Winning strategies in cer …

For a partial answer, let me prove that every strategically closed partial order admits a nearly tactical winning strategy, one that depends only on the previous two moves, that is, on the previous mo …

The answer is yes, because indeed every partial order admits a well founded cofinal (i.e. dominating) subset. There is no need to consider cardinal characteristics. Theorem. Every partial order admits …