If $M_1 \cup M_2 $ is a hyperplane in $\mathbb{R}^n$ then $M_1 = M_2$.

Question

Let $M_1$ and $M_2$ be affine subspaces of $\mathbb{R}^n$ such that they are not proper subsets of each other. If $M_1 \cup M_2 $ is a hyperplane in $\mathbb{R}^n$ then $M_1 = M_2$. What I did is as follows: $dim(M_1 \cup M_2) = n-1$, so $max\{dim(M_1),dim(M_2)\} = n-1$, wlog we can assume $dim(M_1) = n-1$. But I do not know how to continue. Any help is appreciated in advance.

Answer 1

You can try using the following Lemma:

Lemma 1: If $M_1$ and $M_2$ are affine subset of some real vector space $E$ such that $M_1\cup M_2$ is an affine subset of $E$, then $M_1\subseteq M_2$ or $M_2\subseteq M_1$.

In order to prove this, suppose that $M_1\subseteq M_2$ is false, i.e. there is $x\in M_1\setminus M_2$. Let $y\in M_2$, then for any $\lambda\in\mathbb R$, $\lambda x+(1-\lambda) y\in M_1\cup M_2$ by affinity of $M_1\cup M_2$. But the line $\{ \lambda x+(1-\lambda) y : \lambda\in\mathbb R \}$ can intersect $M_2$ only at $y$, otherwise we would have $x\in M_2$. This means that for instance $z=\frac{x+y}{2}\in M_1$. Note that $\{ \lambda x+(1-\lambda) z : \lambda\in\mathbb R \}$ is the same line as before and it is contained in $M_1$, therefore $y\in M_1$, i.e. $M_1\subseteq M_2$.