Let $m$ be a plane in $\mathbb{R}^{3}$ and let $R_{p}: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ is the reflection with respect to the plane $p$.

Let $X \in \mathbb{R}^{3}$ be a countable subset of $\mathbb{R}^{3}$. I would like to prove that there exists a plane $m$ so that $R_{m}(X) \cap X = \emptyset$.

It looks as if there exists a tricky approach using the Baire's theorem, but i do not see any possible way how i can rigorously start, apart from stating some simple observations.

Are there any hints that might help?

  • $\begingroup$ You can start by showing that there exists a plane that misses $ X $. Fix a point $ P \in \mathbb{R}^{3} \setminus X $, and let $ V \stackrel{\text{df}}{=} \left\{ \overrightarrow{P Q} ~ \middle| ~ Q \in X \right\} $. Next, pick a point $ P' \in \mathbb{R}^{3} \setminus \{ P \} $ so that $ \overrightarrow{P P'} $ is not parallel to any vector in $ V $. We can do this as $ V $ is a countable set. Finally, pick a point $ P'' \in \mathbb{R}^{3} $ so that $ \overrightarrow{P P''} $ is not spanned by $ \overrightarrow{P P'} $ and any vector in $ V $. We can do this by the Baire Category Theorem. $\endgroup$ – Berrick Caleb Fillmore Nov 1 '16 at 10:12
  • 1
    $\begingroup$ The plane containing $ P $ and spanned by $ \overrightarrow{P P'} $ and $ \overrightarrow{P P''} $ must then miss $ X $. To solve the problem in its entirety, try to find an argument that employs the Baire Category Theorem in essentially the same manner. $\endgroup$ – Berrick Caleb Fillmore Nov 1 '16 at 10:17

I don't think we need Baire. Let $P$ be the $x$-$y$ plane. Write the points of $X$ as $\{(x_n,y_n,z_n): n \in \mathbb N\}.$ Let $u=(0,0,1).$ The desired plane will be $P+tu$ for some $t\in \mathbb R.$

Clearly we need to avoid any $t \in \{z_n: n\in \mathbb N\}.$ We also need to avoid the situation of two distinct points in $X,$ neither in $P+tu,$ that are symmetric with respect to $P+tu.$ In the latter case we will have $t=(z_m+z_n)/2$ for some pair $m,n.$ These are the only examples of a $P+tu$ that doesn't work.

It follows that if

$$t\not \in \{z_n: n \in \mathbb N\} \cup \{(z_m+z_n)/2 : m,n\in \mathbb N\},$$

a countable set, then $P+tu$ has the desired property. There are of course uncountably many such $t.$

| cite | improve this answer | |

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.