Does it exist a subset $S$ of $\mathbb{R}^4$ such that for all affine hyperplane $H\subset \mathbb{R}^4$, the set $H \cap S$ is dense in $H$ and does not contains $4$ coplanar points?

More than a solution I would like some intermediate questions to solve the problem on my own. Sometimes (for me) exercise are too difficult to be solved without intermediate questions. Yet which seems really interesting.

The problem here is that I cannot visualize what $H \cap S$ looks like. If I change this exercise as to find a dense subset of $\Bbb{R}^3$ wich does not contains $4$ coplanar points is more easier. I can take any countable dense subset $\{x_n\}$ and working by induction. Also the case uncountable is, I think, much more difficult to solve (I do not know if it has a solution but this issue appears naturally).

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    $\begingroup$ If $H\cap S$ is dense in $H$, then $H\cap S$ has at least 4 points, and they are coplanar, because they lie in $H$, aren't they? $\endgroup$ – user63660 Dec 24 '14 at 17:28
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    $\begingroup$ @user63660: $H$ has dimension $3$, coplanarity would indicate a common affine space of dimension $2$. So no, $4$ points in $H$ need not be coplanar. $\endgroup$ – MvG Dec 24 '14 at 20:05
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    $\begingroup$ Are you familiar with transfinite induction/recursion? That’s the easiest way to do it, but if you’re not, I can try to recast it as a Zorn’s lemma argument. $\endgroup$ – Brian M. Scott Dec 24 '14 at 21:13
  • $\begingroup$ @BrianM.Scott unfortunately I am not, it will very nice of you if you can try as Zorn's lemma argument (wich I am 'more' familiar with). $\endgroup$ – user169373 Dec 24 '14 at 22:23
  • $\begingroup$ @MarcGato: Okay; I’ll have to think a bit to make the translation, and tonight is a little busy, so I may not actually get to post it until tomorrow. $\endgroup$ – Brian M. Scott Dec 24 '14 at 22:41

It turns out that I was a over-optimistic. I don’t see a way to do it with Zorn’s lemma that doesn’t get unbearably complicated, so I’m going to go ahead and use transfinite recursion to construct $S$. I’ll do my best to make the explanation self-contained; in particular, I’ll avoid transfinite ordinals entirely. Fortunately, this particular recursion is quite straightforward and greatly resembles more familiar recursive constructions of ordinary sequences of things. Also, I’ll write it up for the $\Bbb R^3$ version of the problem: the mathematics is basically the same, but you can visualize what’s going on much better.

There is a set $S\subseteq\Bbb R^3$ such that $H\cap P$ is dense in $P$ for each plane $P$ in $\Bbb R^3$, but $H$ does not contain $3$ collinear points.

The first key observation is that there are $|\Bbb R|=\mathfrak{c}$ planes in $\Bbb R^3$. Thus, we can enumerate them as $\{P_\xi:\xi\in\Bbb R\}$. We will also need the fact that (assuming the axiom of choice) there is a well-ordering, $\preceq$, of $\Bbb R$ such that $|\{\xi\in\Bbb R:\xi\preceq\eta\}|<\mathfrak{c}$ for each $\eta\in\Bbb R$. We may further assume that $0$ is the first element in this well-ordering, i.e., that $0\preceq\xi$ for all $\xi\in\Bbb R$. (This is just a convenience, to make it easy to talk about the first element.)

Next, for each $\xi\in\Bbb R$ let $\mathscr{B}_\xi=\{B_{\xi,n}:n\in\Bbb N\}$ be a countable base for the relative topology of the plane $P_\xi$.

Finally, for each $A\subseteq\Bbb R^3$ let $L(A)$ be the union of the lines determined by pairs of points of $A$.

Suppose that $\eta\in\Bbb R$, and for each $\xi\prec\eta$ we have constructed a set $S_\xi\subseteq\Bbb R$ such that:

  • $S_\xi\setminus\bigcup_{\zeta\prec\xi}S_\zeta$ is countable;
  • no three distinct points of $S_\xi$ are collinear;
  • $S_\zeta\subseteq S_\xi$ for all $\zeta\preceq\xi$; and
  • $S_\xi\cap P_\xi$ is dense in $P_\xi$.

This is the recursion hypothesis; we’ll use it to construct a set $S_\eta$ that satisfies all four conditions with $\xi$ replaced by $\eta$. Note that it is vacuously satisfied when $\eta=0$, so the basis step will be no different from any later step.

Now I’ll construct $S_\eta$.

Let $T_0=\bigcup_{\xi\prec\eta}S_\xi$. Let $x,y$, and $z$ be distinct points of $T_0$. Then there are $\zeta,\theta,\xi\prec\eta$ such that $x\in S_\zeta$, $y\in S_\theta$, and $z\in S_\xi$. Without loss of generality $\zeta\preceq\theta\preceq\xi$, so $x,y,z\in S_\xi$ and by the recursion hypothesis are not collinear. Thus, no three distinct points of $T_0$ are collinear. If $T_0\cap P_\eta$ is dense in $P_\eta$, let $S_\eta=T_0$; clearly this satisfies the recursion hypothesis with $\xi$ replaced by $\eta$.

Now assume that $T\cap P_\eta$ is not dense in $P_\eta$. $T_0$ is the union of fewer than $\mathfrak{c}$ countable sets, which means that it’s the union of countably many sets, each of cardinality less than $\mathfrak{c}$. $\Bbb R$ is not the union of countably many sets of smaller cardinality, so $|T_0|<\mathfrak{c}$. (Here I’m using a basic fact about infinite cardinal arithmetic; I don’t see a way to avoid it.) It follows that pairs of points of $T_0$ generate fewer than $\mathfrak{c}$ lines, and hence that $L(T_0)\cap P_\eta$ is the union of fewer than $\mathfrak{c}$ points and lines.

We’ll now do an ordinary recursion over $\Bbb N$ to expand $T_0$ to the desired set $S_\eta$.

Suppose that $n\in\Bbb N$, and for each $k<n$ we’ve chosen a point $x_{\eta,k}\in B_{\eta,k}$ in such a way that no three points of $T_n=T_0\cup\{x_{\eta,k}:k<n\}$ are collinear. If $T_n\cap B_{\eta,n}\ne\varnothing$, let $x_{\eta,n}$ be any point of $T_n\cap B_{\eta,n}$. Otherwise, fix any point $p\in B_{\eta,n}$. $L(T_n)\cap P_\eta$ is the union of fewer than $\mathfrak{c}$ points and lines, and there are $\mathfrak{c}$ lines through $p$, so there is a line $L_0$ in $P_\eta$ through $p$ that intersects each line in $L(T_n)$ in at most one point. Thus, $|L_0\cap L(T_n)|<\mathfrak{c}$, while $|L_0\cap B_{\eta,n}|=\mathfrak{c}$, so we may choose a point $x_{\eta,n}\in (L_0\cap B_{\eta,n})\setminus L(T_n)$. Clearly no three distinct points of $T_{n+1}=T_n\cup\{x_{\eta,n}\}$ are collinear, so the recursive construction goes through. Then we set $S_\eta=\bigcup_{n\in\Bbb N}T_n=T_0\cup\{x_{\eta,n}:n\in\Bbb N\}$ and see that all four conditions of the main recursion are satisfied with $\xi$ replaced by $\eta$.

It’s a theorem of set theory that we can now conclude that we have family of set $\{S_\eta:\eta\in\Bbb R\}$ that satisfy the recursion hypothesis for each $\eta\in\Bbb R$. This is entirely analogous to the ordinary recursive construction of the points $x_{\eta,n}$ for each $n\in\Bbb N$, at the end of which we concluded that we had the whole set of them, $\{x_{\eta,n}:n\in\Bbb N\}$, to be lumped in with $T_0$ to form the set $S_\eta$.

Now let $S=\bigcup_{\xi\in\Bbb R}S_\xi$. The verification that no three distinct points of $S$ are collinear is essentially the same as the verification of the corresponding fact for $T_0$ above, and the fourth clause of the recursion hypothesis ensures that $S$ has dense intersection with each plane in $\Bbb R^3$.

  • $\begingroup$ It will take me some time to go into details about set theory but I think I understand the underlying idea. Anyway it was a pleasure to read this answer, fulfilling and well explained. $\endgroup$ – user169373 Dec 26 '14 at 11:19
  • $\begingroup$ See also mathoverflow.net/questions/93601/… for other examples of application of this method. What is unclear, though, if such construction can be done without using some form of the Axiom of Choice. $\endgroup$ – Moishe Kohan Dec 26 '14 at 18:43
  • $\begingroup$ @MarcGato: If after thinking about it you have any questions, please do feel free to ask. $\endgroup$ – Brian M. Scott Dec 26 '14 at 19:30
  • $\begingroup$ As I am on holidays I took my day to think about it and I think I completly understandoot it. Thanks. $\endgroup$ – user169373 Dec 26 '14 at 19:37
  • $\begingroup$ @MarcGato: You’re very welcome. $\endgroup$ – Brian M. Scott Dec 26 '14 at 23:20

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