Intersection of affine subspaces is affine If I have two affine subspaces, each is a translation (or coset) of some linear subspace.  I want to show that the intersection of such affine subspaces is also affine, particularly in $\mathbb{R}^d$.  My intuition suggests that the resulting space is just a coset of the intersection of the two linear subspaces, but I'm having some trouble arguing this precisely.
 A: Hint: Let $A_1 = x_1 + U_1$, $A_2 = x_2 + U_2$ your two affine subspaces, if $A_1 \cap A_2 = \emptyset$, we are done, otherwise there is an $x \in A_1 \cap A_2$. But then $A_1 = x+ U_1$ and $A_2 = x+ U_2$ ... does this help?
A: As an exercise you should be able to verify the following.

Theorem. Let $V$ be a linear space, $U \subseteq V$ be a linear subspace, and $x, \, y \, \in V$ be any two elements. Then the following are equivalent

*

*$x - y \in U$

*$x + U = y + U$

*$(x + U) \cap (y + U) \neq \emptyset$

Now, suppose that $A_1 = x_1 + U_1$ and $A_2 = x_2 + U_2$ are two affine subspaces. Assume $A_1 \cap A_2$ is not empty and choose any $y \in A_1 \cap A_2$. $y \in A_1$ implies that $y = x_1 + u_1$ for some $u_1 \in U_1$, which leads to $y - x_1 \in U_1$. Using the above theorem, this leaves us with $A_1 = y + U_1$. Similarly, you can show that $A_2 = y + U_2$. Again, suppose $z \in A_1 \cap A_2$. This implies that $z = y + u_1 = y + u_2$. This in turn requires that $u_1 = u_2 = u \in U_1 \cap U_2$. Consequently, $z = y + u$ for some $u \in U_1 \cap U_2$. Thus $z \in y + (U_1 \cap U_2)$, implying that $A_1 \cap A_2 \subseteq y + (U_1 \cap U_2)$. Proving the inclusion in the other direction is straight forward and yields the final result as
$$A_1 \cap A_2 = (x_1 + U_1) \cap (x_2 + U_2) = (y + U_1) \cap (y + U_2) = y + (U_1 \cap U_2)$$
for some $y \in (x_1 + U_1) \cap (x_2 + U_2)$.
