Dimension of the sum of three subspaces We know that $$\dim(U_1 + U_2) = \dim U_1 + \dim U_2 - \dim(U_1 \cap U_2)$$ if $U_1$ and $U_2$ are finite dimensional subspaces.
For three finite dimensional subspaces prove or give a counterexample for the following:
$$
\begin{align}
\dim(U_1 + U_2 + U_3) &= \dim U_1 + \dim U_2 + \dim U_3 \\
&- \dim(U_1 \cap U_2) - \dim(U_2 \cap U_3) - \dim(U_1 \cap U_2)\\
&+ \dim(U_1 \cap U_2 \cap U_3) 
\end{align}$$
Its basically the formula for the union of three sets. I think that this is false but the only reason I can think it would be is because the union of subspaces is rarely a subspace itself, even though it applies to sets, so it seems to me like this formula which works for sets shouldn't work for subsets because it would disrupt overlapping elements of $U_1 + U_2 + U_3$ and run into complications with the sum being closed under addition or something like that. Help much appreciated thank you!

This is the counterexample given by Willie Wong on a mathoverflow post with this question:
Consider $3$ distinct lines through the origin in $\mathbb{R^2}$ as $U_1$, $U_2$, and $U_3$.
Then the left hand side of the equation becomes
$$\mathrm{Dim}(U_1+U_2+U_3)=\mathrm{Dim}(\mathbb{R^2})=2,$$ while the right hand side of the equation becomes
$$\mathrm{Dim}(U_1)+\mathrm{Dim}(U_2)+\mathrm{Dim}(U_3)-\mathrm{Dim}(U_1\cap U_2)-\mathrm{Dim}(U_1\cap U_3)-\mathrm{Dim}(U_2\cap U_3)+\mathrm{Dim}(U_1\cap U_2\cap U_3)\\=3\mathrm{Dim}(\mathbb{R})-3\mathrm{Dim}(\{0\})+\mathrm{Dim}(\{0\})=3,$$
and as $2\ne3$ this equation does not hold.
