Construct a function that will be disjoint with continuum many lines

Question

The symbol $\mathbb L$ will stand for the family of all lines in the plane that are neither horizontal nor vertical. Also, we put $\mathbb L_0:=\{\ell\in\mathbb L\colon \ell(0)=0\}$

lemma. Let $\mathcal F:=\{f_\eta\colon \eta<\mathfrak c\}$ be a family of distinct functions from $\Bbb R$ to $\Bbb R$ such all functions will be repeated continuum many times. Then there exists a function $h\colon \Bbb R\to \Bbb R$ such that for $f\in\mathcal F$ the family $$G_f:=\{\ell\in\mathbb L_0\colon (h+f)(x)\neq \ell(x) \, \text{for every nonzero}\, x\in\Bbb R\},$$ has a cardinality continuum.

Proof. Let $\{x_{\eta}\colon \eta<\mathfrak c\}$ be an enumeration of the set of real numbers $\Bbb R$. By transfinite induction on $\eta<\mathfrak c$ we construct a sequence\ $\langle\langle h_\eta,D_\eta,\ell_\eta\rangle\,\colon\eta<\mathfrak c\rangle$ satisfying the following inductive conditions for every $\eta<\mathfrak c$

(a)$D_\eta\in[\Bbb R]^{<\mathfrak c}$, $h_\eta\colon D_\eta\to\Bbb R$, and $\ell_{\eta}\in\mathbb L_0;$

(b) $x_\eta\in D_\eta$, $h_{\zeta}\subset h_{\eta}$ for every $\zeta\leq\eta;$

(c) $h_{\eta}(x)+f_{\zeta}(x)\neq \ell_{\zeta}(x)$ for all $\zeta<\eta$ and $x\in D_{\eta}\setminus D_\zeta$;

(d) $h_{\eta}(x)+f_{\eta}(x)\neq\ell_{\eta}(x)$ for every nonzero $x\in D_{\eta}.$

If such a sequence can be found, then $h:=\bigcup_{\eta<\mathfrak c}h_\eta$ is as desired. Indeed, $h\in\Bbb R^\Bbb R$ by (a), (b). if $\zeta<\mathfrak c$ is such that $f_\zeta=f,$ Indeed, for $x\in D_\zeta$ we have $(h+f)(x)=h_\zeta(x)+f_\zeta(x)\neq \ell_\zeta(x)$ is ensured by (f) used with $\eta=\zeta$. At the same time, if $x\in \Bbb R\setminus D_\zeta$, then there exists an $\eta<\mathfrak c$ such that $x\in D_{\eta}\setminus D_\zeta$ and the condition (d) implies that $(h+f)(x)=h_\eta(x)+f_\zeta(x)\neq \ell_\zeta(x)$. Since each $f$ will appear continuum many time so this shows $G_f$ has cardinality continuum.

My question is :

Is the above argument mathematical correct?

Answer 1

This is not real an answer I just want to say your solution is correct.

Answer 2

According to Gob your solution is correct, which I would agree with: it seems like if there were some $\zeta,\eta$ such that $f_\zeta+h_\eta$ is not disjoint from $\ell_\eta$ on its domain, we could pick an ordinal $\xi=\textrm{max}\{\zeta,\eta\}$ and consider the step in which $(h_\xi,D_\xi,\ell_\xi)$ is defined, contradicting the inductive definition.

Two things about this solution:

• How do we know that from each step of the inductive definition we can proceed without revising an earlier function (clause (b))?
• Additionally, I believe some parts can be simplified: for example, clause (d) in the $\alpha$th step of the induction is implied by clause (c) in the $\alpha+1$th step, since we may set $\eta=\zeta<\eta+1$, and $h_{\alpha+1}(x)=h_\alpha(x)$ for any $x$ in $h_\alpha$'s domain, we get $h_\alpha(x)+f_\alpha(x)=h_{\alpha+1}(x)+f_\alpha(x)\neq\ell_\alpha(x)$.