Let $U,V,W$ be three vector subspaces of a same vector space.
It is well-known that $$\dim (U + V) = \dim U + \dim V - \dim (U \cap V)$$ works but $$ \dim(U +V + W) = \dim U + \dim V + \dim W - \dim (U \cap V) - \dim (U \cap W) - \dim (V \cap W) + \dim(U \cap V \cap W) $$ fails.
The counterexample is given by three distinct lines in $\mathbb{R}^2$.
Questions:
- (why failed) Philosophically vector spaces are completely determined by its basis. We have in-ex principle for three sets, in particular, the set of basis. Why the same in-ex principle fails for three vector spaces? (I notice some post and comments here but the failure of distributive law did not solve my doubt.)
- (how to fix) It is well-known that ({vector subspaces of $K$},$+$,$\cap$) forms a (bounded modular) lattice (c.f. wiki). ({subsets of $S$},$\cup$,$\cap$) is also a lattice. If I take $S$ to be a basis for $K$, then we get a free-forgetful adjunction pair that is lattice isomorphism. To be precise, $\langle S_1+S_2\rangle=\langle S_1\rangle+\langle S_2\rangle$, $\langle S_1\cap S_2\rangle=\langle S_1\rangle\cap \langle S_2\rangle$, where the angle bracket is the free span. In this way, I get isomorphic lattices. Then I suppose this is the correct form for the inclusion-excluision principle, i.e., under the condition that we have to fix a basis $S$ and $U,V,W$ must be the span of some subset of $S$? Any better form of in-ex principle for vector spaces is also welcome.