$F=\bar{F}$, $int F = \varnothing$, $E = \{(x,x^2)\}$. There is $(a,b)\in E$ that for any $q\in Q^+$, $c\in F$: $b \neq q(a-c)$

Question

Let $F$ be closed set, $F\subseteq \mathbb{R}$ and $int F = \varnothing$. Let $E = \{(x,x^2): x \in \mathbb{R} \} \subseteq \mathbb{R}^2$. Prove that there is point $(a,b)\in E$ that for any $q\in \mathbb{Q}, a > 0$ and $c\in F$: $b \neq q(a-c)$

Ok, so the obvious thing is that one has to use Baire theorem to prove it however I have some problem with the construction of that bounded sets with empty interior.

So let define "bad" points as those which do hold equality $b = q(a-c)$. We remember that we are on parabola so the equality is $a^2 = q(a-c)$. Let $W_q = \{(a,a^2) \in E: a^2 = q(a-c) $ for some $q\in \mathbb{Q}^+$$ \}$

It's obvious that $W = \bigcup W_q$ is the set of all bad points and if we prove that $W_q$ are closed and $int W_q = \emptyset$ then the prove is completed because by Baire theorem that would mean that there exist $a \in \mathbb{R}$ that $(a,a^2) \notin W$

And here the troubles starts.

At first, I thought that maybe $|F| <= |\mathbb{N}|$ but there are sets that have empty interior and are closed but still have many elements (for example Cantor's set) so it is the wrong way.

So maybe $W_q$ are homeomorphic to $F$ and that would implicate that $W_q$ are closed and have empty interiors, but since $|F|$ may not be equal to $|W_q|$ it is also not true.

So how can I show that $W_q$ are closed sets and $int W_q = \emptyset$?

Answer 1

Your reasoning is well.

First notice that if we define $f(x)=x^2$, $f$ is continuous and we have that $E=\mathrm{graph}(f)$ so $E$ is close. Besides given $q$ we can define $g_q(y,z)=z-q(y-c)$, where $g_q$ is continuous (so pre-image of close is close) so as $W_q=g_q^{-1}(\lbrace 0\rbrace)\cap E$, $W_q$ is close also.

Now notice that $\mathrm{int}(E)=\emptyset$ and, as $W_q\subset E$, then $\mathrm{int}(W_q)=\emptyset$