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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?

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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 $t.$

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