Let $\mathcal F:=\{f_\xi\colon \xi<\mathfrak c\}$ be a family of functions from $\Bbb R$ to $\Bbb R$ and $K$ be a nonempty perfect set. The question is
Construct a function $g\colon \Bbb R \to \Bbb R$ such that $(g+f)[P]\subsetneq K$ for any $f\in \mathcal F$ and for any nonempty prefect set $P$.
Here is my attempt: let $\mathcal P:=\{P_\xi\colon \xi<\mathfrak c\}$ be the set of all prefect subsets of $\Bbb R.$ Let $B$ be a Bernstein subset of $\Bbb R$ and define, by induction on $\xi<\mathfrak c$, a sequence $$ x_\xi\in B\cap P_\xi\setminus\{x_\zeta\colon \zeta<\xi\}. $$ Put $D:=\{x_\xi\colon \xi<\mathfrak c\}$. Fix a bijection function $\Phi\colon \mathcal P \times K \times\mathcal F\to D$. By transfinite induction on $\eta<\mathfrak c,$ we construct a sequence $\langle\langle g_\eta,D_\eta\rangle\colon\eta<\mathfrak c\rangle$ satisfying the following inductive conditions for every $\eta<\mathfrak c$:
(a) $g_\eta\colon D_\eta\to\Bbb R$ and $g_{\zeta}\subset g_{\eta}$ for every $\zeta\leq\eta$;
(b) If $x\in\Phi(P,K,f)\cap D_{\eta}$ then $(g_{\eta}+f)(x)\not\in K$.
If such a sequence can be found, then $g:=\bigcup_{\eta<\mathfrak c}g_\eta$ is as needed. Clearly, $g$ is a function. To see $g$ has the desired property. Fix a prefect set $P$ and $f\in\mathcal F$, there exist $\xi<\mathfrak c$ such that $\langle P, K, f\rangle= \langle P_\xi, K_\xi, f_\xi\rangle$ and $x_\xi\in D\cap P_\xi$ such that, by (b), $(g+f_\xi)(x_\xi)\notin P_{\xi}$. Then, $(g+f)[P]\subsetneq K$.
Could you please read this short argument and tell me if you see any mistake?
