0
$\begingroup$

Let $F\subset\mathbb{R}$ be a closed nowhere dense set. One must show there exists $(a,b)\in S^1$ for which $b\neq qa+c$, for all $q\in\mathbb{Q},c\in F$.

It's my second question concerning Baire category theorem - I already know what the crux is -- to find a family of closed nowhere dense sets, preferably (in this example) connected with $q,c$ and thus with the set $F$. But no such family comes to my mind. I suppose though, it should comprise some subsets of $F$. I wanted to thoroughly think about it, so what I need is just a little hint.

$\endgroup$
  • $\begingroup$ You write "$q,c \in F$ are rationals". But from the assumptions on $F$ I don't see that $F$ necessarily contains any rationals. How does this go together? $\endgroup$ – jflipp Nov 24 '14 at 13:42
  • $\begingroup$ Oh, I'm very sorry, I got carried away. Thanks to the editor, it should be like it is now. $\endgroup$ – Jules Nov 24 '14 at 14:01
1
$\begingroup$

Suppose , that assertion is not true. This means that for every $(a,b)\in\{(x,y)\in\mathbb{R}^2 :x^2 +y^2 =1\}=S^1$ there exists a rational number $q\in\mathbb{Q}$ and $c\in F$ such that $b=qa+c$. For a given rational number $q\in \mathbb{Q}$ denote by $A_q =\{(u,v)\in S^1 :v-qu\in F\} .$ Since $S^1 = \bigcup_{q\in\mathbb{Q}} A_q$ and $S^1$ is complete metric space ( with induced from $\mathbb{R}^2$ Euclidean metric ) thus from Baire'a Theorem there exists $q_0 \in \mathbb{Q}$ such that the set $A_{q_0}$ is dense in some ball on the circle $S^1 .$ But this means that the set $\overline{A_{q_0}}$ contain some open arc of this circle. Let us now define a map $f:S^1 \rightarrow \mathbb{R}$ by $f(x,y) =y-q_0 \cdot x .$ Since $f$ is continuous $f(\overline{A_{q_0}} ) \subset \overline{f(A_{q_0} )} \subset \overline{F} =F$ but the last shows that the set $F$ contains a connectet subset with two or more elements, so $F$ must contains an open interval. Contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.