Union of two affine subspaces

Question

I had to find a condition under which the following holds:

Let $F_1$ and $F_2$ be affine subspaces of an affine space $\epsilon$. $F_1$∪ $F_2$ is an affine subspace.

Could somebody please help me? Thank you!

Answer 1

Let $\mathbb A$ be an affine space on a field of characteristic $\neq 2$, and let $F_1, F_2$ be two affine subspaces of $\mathbb A$ such that $F_1\,\cup\,F_2$ is still an affine subspace of $\mathbb A$. We want to prove that in this hypothesis we have

$$ F_1 \subseteq F_2 \text{$\;\;\;$or$\;\;\;$} F_2 \subseteq F_1. $$

If we suppose that our thesis is false, we can take $P\in F_1-F_2\,$ and $\,Q\in F_2-F_1$. Since $F_1\cup F_2$ is an affine subspace by hypothesis, we can consider the point $\,R\,$ belonging to that union, given, in barycentric coordinates (see Note 2 below), by

$$ R = \alpha P + \beta Q \hskip8ex [\alpha+\beta=1]$$

for fixed $\alpha,\beta$ (one can take for example $\alpha=\beta=1/2$, thus obtaining the midpoint of the segment $PQ$). Since $R\in F_1\cup F_2$, $R$ belong to $F_1$ or to $F_2$, say the first for example (the other case is analogous). It follows that, if $\lambda+\mu=1$, we have

$$ \lambda R + \mu P \in F_1 $$

what leads to a contradiction by choosing (if $\beta\neq 0$, how certainly we can suppose)

$$ \hskip10ex\lambda = \frac 1 \beta \qquad \mu = -\frac{\alpha}{\beta} \hskip12ex [\text{Note that}\;\;\lambda+\mu=\frac 1 \beta-\frac{\alpha}{\beta}=\frac{1-\alpha}{\beta}=\frac \beta \beta=1] $$

thus obtaining

$$ \frac 1 \beta R -\frac{\alpha}{\beta} P = \frac 1 \beta (\alpha P + \beta Q) -\frac{\alpha}{\beta} P = Q \in F_1, $$

which contradicts the assumption $Q\in F_2-F_1$.

From this contradiction we obtain our thesis.

If characteristic is 2 theorem fails. Example: $\mathbb A = \mathbb Z_2$ over the field $\mathbb Z_2$. In this case, $F_1=\{0\}$ and $F_2=\{1\}$ are two affine subspaces such that neither $F_1 \subseteq F_2$ nor $F_2 \subseteq F_1$, and nevertheless $F_1 \cup F_2 = \mathbb Z_2$ is an affine space.

Note 1 The converse of statement is obvious: if, for example, $F_1 \subseteq F_2$, it follows $F_1\cup F_2= F_2$, an affine subspace.

Note 2 In general, we denote by $\alpha P + \beta Q$ the point

$$ R = O + \alpha(P-O) + \beta(Q-O), $$

where $O$ is any fixed point of the affine space. This is possible because $R$ is independent of $O$ whenever $\alpha+\beta=1$, as it is easy to prove.