Problem
You might guess, by analogy with the formula for the number of elements in the union of three subsets of a finite set, that if $U_1, U_2, U_3$ are subspaces of a finite-dimensional vector space, then $$\dim(U_1 + U_2 + U_3) = \dim U_1 + \dim U_2 + \dim U_3$$ $$- \dim (U_1 \cap U_2) - \dim(U_1 \cap U_3) - \dim(U_2 \cap U_3)$$ $$+ \dim(U_1 \cap U_2 \cap U_3)$$ Prove this or give a counterexample.
Solution
Let $V = \mathbb{R}^3$ and $U_1 = \textrm{span}((1,0,0))$, $U_2 = \textrm{span}((0,1,0))$ and $U_3 = \textrm{span}((1,1,0))$. Then: $$2 = \dim (U_1 + U_2 + U_3) = \dim U_1 + \dim U_2 + \dim U_3 - \dim (U_1 \cap U_2) - \dim(U_1 \cap U_3)$$ $$- \dim(U_2 \cap U_3) + \dim(U_1 \cap U_2 \cap U_3) = 1 + 1 + 1 - 0 - 0 - 0 + 0 = 3$$ A contradiction.