Independent families on $\omega$ with an additional splitting property

Question

Let $\mathscr{F}$ be an independent family, that is, a collection of subsets of $\omega:=\{0,1,\ldots\}$ such that, for all integers $n\ge 1$ and all distinct $X_1,\ldots,X_n \in \mathscr{F}$, then $$ \forall (e_1,\ldots,e_n) \in \{0,1\}^n,\quad X_1^{e_1} \cap \cdots \cap X_n^{e_n} \,\,\text{ is infinite} $$ where $X^0:=X$ and $X^1:=\omega\setminus X$ for all $X\subseteq \omega$.

It is well known that independent families of cardinality $\mathfrak{c}$ exist, see e.g. here for eight different proofs of this fact (Theorem 3.4). Independent families with additional properties (e.g., selective independent families) have been studied, e.g., here and here. Here we ask about a variant of these properties:

Question. Let $\Omega$ be the collection of all $X\subseteq \omega$ which are neither finite nor cofinite. Does there exist an independent family $\mathscr{F}$ of cardinality $\mathfrak{c}$ and a bijection $f: \Omega \to \mathscr{F}$ such that $$ \forall X\in \Omega, \quad X^0 \cap f(X) \,\,\text{ and }\,\, X^1 \cap f(X)\,\,\text{ are both infinite}\,? $$

It is clear that if $\mathscr{F}$ is an arbitrary independent family of cardinality $\mathfrak{c}$ and $f: \Omega \to \mathscr{F}$ an arbitrary bjection, then the answer is negative: indeed, if $f(X)\subseteq X$ for some $X\in \Omega$, then $X^1 \cap f(X)$ would be empty.

Answer 1

Edit: My previous answer had a mistake. Here is an updated answer which still contains some hopefully useful information:

Assuming $\operatorname{cov}(\mathcal{M}) = \mathfrak{d}$, there is a positive answer.

Recall that $\mathfrak{d}= \kappa$ is least such that there is a sequence $\langle K_\alpha : \alpha < \kappa\rangle$ of compact subsets of $2^\omega$ such that $\bigcup_{\alpha < \kappa} K_\alpha = \Omega$.

Consider the partial order $\mathbb{P}$ consisting of monotone functions $h \colon 2^{\leq n} \to 2^{<\omega}$, for some $n \in \omega$, ordered by extension. If $G$ is $\mathbb{P}$-generic over $M$ and $H := \bigcup_{h \in G} h$, then $f \colon 2^\omega \to 2^\omega$ such that $f(x) = \bigcup_{n \in \omega} H(x\restriction n)$ is a continuous bijection between $2^\omega$ and $f''2^\omega$.

We will recursively construct continuous functions $f_\alpha \colon K_\alpha \to \Omega$ for $\alpha < \kappa$. Namely, given $\langle f_\beta: \beta < \alpha \rangle$ for $\alpha < \kappa$, let $M_\alpha$ be an elementary submodel of some large $H(\theta)$ of size $\vert \alpha \vert < \kappa$ containing all $f_\beta$ for $\beta < \alpha$ and $K_\alpha$. Force with $\mathbb{P}$ over $M_\alpha$ and get a generic continuous function $f \colon 2^\omega \to 2^\omega$. This is possible since $\mathbb{P}$ is a countable poset and $\alpha < \operatorname{cov}(\mathcal{M})$. Let $f_\alpha := f \restriction K_\alpha$.

Finally, let $F \colon \Omega \to \Omega$ be such that $F(x) = f_\alpha(x)$ for $\alpha$ least such that $x \in K_\alpha$. To see that $F''\Omega$ is independent we note that:

Claim Let $c$ be Cohen over $M$ and $B \in M$ an analytic independent family. Then $B \cup \{ c \}$ is independent.

This is Lemma 5.8 here and is essentially due to an argument of Arnie Miller that can be found in here.

Since $f_\alpha(x_0), \dots, f_\alpha(x_k)$ are mutually Cohen generic over $M_\alpha$, any Boolean combination is Cohen as well and can be added to any finite union of the compact sets $f_\beta ''K_\beta \in M$, for $\beta < \alpha$, to form an independent set. Thus also the union of all $f_\beta '' K_\beta$, $\beta \leq \alpha$ is independent.

Finally, to see that $x \in K_\alpha$ hits and avoids $f_\alpha(x)$ infinitely, make the following genericity argument: Let $h \in \mathbb{P}$ be arbitrary, say the domain of $h$ is $2^{\leq n}$ and the range is contained in $2^{\leq m}$. By compactness, there is $k > \max(m,n)$ such that for any $x \in K_\alpha$, $x(i) = 1$ for some $i \in (\max(m,n),k]$. We can simply extend $h$ to $h'$ of domain $2^{\leq k}$ such that $h'(s)$ hits $s$ above $\max(m,n)$ for every $s \in 2^{\leq k}$. Similarly for avoiding instead of hitting. Since $K_\alpha \in M_\alpha$, this genericity argument can be made over $M_\alpha$.