Prove that there is a plane in $\mathbb{R}^n$ with the following property

Question

Let $X \subset \mathbb{R}^n$ be a union of countably many affine subspaces of $\mathbb{R}^n$ with dimension less or equal $n-2$.

Prove that $\mathbb{R}^n \setminus X$ is path connected.

Assume $n = 2$. Then we need to show, that the plane with countably many points removed is path connected.

Take $x_1, x_2 \in \mathbb{R}^2 \setminus X$. We find two nonparallel lines in $\mathbb{R}^2$ going through $x_1$ and $x_2$ which do not cross $X$. It can be done because the cardinality of the set of all lines through $x_i$ is $|\mathbb{R}|$ and the cardinality of $X$ is just $|\mathbb{N}|$. These two lines intersect at some point and we get a path between $x_1$ and $x_2$ in $\mathbb{R}^2 \setminus X$.

Now I want to reduce the initial problem to the case with $n=2$. In order to do so I need to

prove that for any two points $x_1$ and $x_2$ in $\mathbb{R}^n \setminus X$ there is a plane in $\mathbb{R}^n$ containing both these points and intersecting $X$ in at most countably many points.

How can I do that?

Answer 1

Let $\{ A_k \}_{k = 1}^\infty$ be affine subspaces of $\mathbb{R}^n$ of dimension $n-2$, and take distinct points $x, y \in \mathbb{R}^n$. Translating, one can assume $x = 0$ and $0 \notin A_k \forall k \in \mathbb{N}$.

There is a homeomorphism $\Phi: \mathcal{G} \rightarrow \mathbb{P}^{n-2} ( \mathbb{R} )$ where $\mathcal{G}$ is the space of $n-2$ dimensional affine spaces intersecting $\mathbb{R}y$. We construct this map as follows. Let $U = \mathbb{R} y$ and $V = U^{\perp}$. $V$ is an $n-1$ dimensional vector space. Take an $n-2$ dimensional vector space $W$ containing $y$. $W$ is uniquely determined by $W \cap V$, since $W = W \cap (V \oplus U) = (W \cap V ) \oplus (W \cap U) = (W \cap V) \oplus U$. There is a vector space isomorphism $T : V \rightarrow \mathbb{R}^{n-1}$. $T(W \cap V)$ is then a subspace of $\mathbb{R}^{n-1}$. Obviously \begin{align*} dim_{\mathbb{R}}(T(W \cap V)) = dim_{\mathbb{R}}( W \cap V) = dim_{\mathbb{R}} ( W ) - dim_{\mathbb{R}} ( U ) = n - 2 \end{align*} Thus $T(W \cap V)^{\perp}$ is a line through the origin of $\mathbb{R}^{n-1}$. Define $\Phi (W) = T(W \cap V)^{\perp}$.

The following is rather straightforward, so I leave it as an exercise:

Exercise: For each $k \in \mathbb{Z}$ the set $\mathcal{G}_k = \{ z \in \mathbb{P}^2 ( \mathbb{R} ) : |\Phi^{-1}(z) \cap A_k| \leq 1 \} $ is open and dense in $\mathbb{P}^2 (\mathbb{R})$.

The claim follows therefore from the following lemma:

Lemma: Baire Category Theorem Let $\mathcal{X}$ be a locally compact Hausdorff space and take dense open sets $\{ U_i \}_{i = 1}^\infty$ in $\mathcal{X}$. Then $\cap_{i = 1}^\infty U_i$ is dense in $\mathcal{X}$.

Spaces with this property are called Baire spaces- spaces in which the intersection of countably many dense sets are dense. See here: https://en.wikipedia.org/wiki/Baire_space

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Edit: here is an elaboration on how to finish the proof from the lemma, as requested: from the exercise we know that $\mathcal{G}_k$ are open and dense in $\mathbb{P}^2 (\mathbb{R})$. Since $\mathbb{P}^2 ( \mathbb{R} )$ is compact and Hausdorff, $\mathcal{H} = \cap_{i = 1}^\infty \mathcal{G}_k$ is dense by the Baire Category Theorem. But this is much stronger than we need. The upshot is that $\mathcal{H}$ is nonempty. Take $z \in \mathcal{H}$. Then $|\Phi^{-1} (z) \cap A_k| \leq 1 \forall k \in \mathbb{N}_{\geq 1}$, so that $\left|\Phi^{-1} (z) \cap \bigcup_{k \in \mathbb{N}_{\geq 0}} A_k \right|$ is countable. $\Phi^{-1} (z)$ is therefore the desired plane.

Answer 2

As is suggested in the comments it is possible to do what you are suggesting, but it will require some measure theory.

In particular you have a $\mathbb{P}^{n-2}(\mathbb{R})$ worth of planes through $x_1$ and $x_2$. For any dimension $d \le n-2$ subspace $V$ of $\mathbb{R}^n$ those planes through $x_1$ and $x_2$ that intersect $V$ in a line cut out a subvariety of $\mathbb{P}^{n-2}(\mathbb{R})$ of strictly smaller dimension (cut out by some determinants vanishing) and in particular this has measure zero. The countable union of measure zero sets has measure zero so in particular we can find a plane through $x_1$ and $x_2$ that meets all the subspaces in $X$ in dimension at most 1, as desired.

I'll note that if you want to avoid measure theory and just use facts about cardinality you can still solve the overall problem, but by a slightly different strategy.

Note that for each dimension $n-2$ subspace $V$ in $X$ there is exactly $1$ hyperplane containing $V$ and $x_1$ (and the same for $x_2$). In particular for cardinality reasons we can find nonparallel hyperplanes through $x_1$ and $x_2$ that meet all the subspaces in $X$ in dimension at most $n-3$.

But now we can apply induction to get that the complement of $X$ in each hyperplane is path connected. So to make a path from $x_1$ to $x_2$ we can just find a path from $x_1$ to some point in the intersection of these two hyperplanes, and then a path from that to $x_2$

Answer 3

Proof for $\mathbb R^3:$ Suppose $L_1,L_2, \dots$ are lines in $\mathbb R^3.$ Assume $p,q\in \mathbb R^3 \setminus (\cup L_k).$ WLOG, we can take $p=(0,0,1),q=(0,0,-1).$

For each $m>0$ let $P_m$ denote the plane spanned by $p$ and $(1,m,0).$ Let $S$ be the unit sphere, and define the unit circle $C_m = P_m\cap S.$ Now each $L_k$ can intersect $S$ in at most two points. It follows that $\cup L_k$ can intersect only countably many $C_m.$ But there are uncountably many $C_m.$ Thus there must exist $m$ such that $(\cup L_k)\cap C_m= \emptyset$ (in fact there are uncountably many such $m$). For any such $m$ we have an entire circle $C_m$ connecting $p$ to $q$ within $\mathbb R^3 \setminus (\cup L_k).$