Positivity of the alternating sum associated to at most five subspaces Let $V_1 , V_2 , \dots , V_n $ be vector subspaces of $ \mathbb{C}^m$ and let  $$\alpha = \sum_{r=1}^n (-1)^{r+1} \sum_{  \ i_1 < i_2 < \cdots < i_r } \dim(V_{i_1} \cap \cdots \cap V_{i_r})$$  
For $n = 2$ we have the equality $ \alpha =  \dim(\sum_{i = 1}^{n} V_i) $; it's false for $n>2$, see this answer.
For $n=3$, we have only the inequality $ \alpha \ge  \dim(\sum_{i = 1}^{n} V_i) $; it's false for $n>3$, see this post.
For $n>5$, the inequality $\alpha  \ge 0$ is false in general, see the comment of Darij Grinberg below.   
Question: Is it true that $\alpha \ge 0$, in the case $n \le 5$?   
Remark: I think this question interesting for itself; it admits also applications in the interaction between representations theory and subgroups lattice.  
 A: Here is a purely combinatorial proof for $n=5$. We first generalize the problem as follows:
Let $B_5$ be the boolean lattice of rank $5$, i.e. the subsets lattice of $\{1,2,3,4,5\}$.
Lemma:  Let $\phi: B_5 \to \mathbb{R}_{\ge 0}$ be a map  satisfying that  $\forall a, b \in B_5$:
$(1)$ $ \ $ $a \le b \Rightarrow \phi(a) \le  \phi(b)$   [poset morphism]
$(2)$ $ \ $ $\phi(a \vee b) + \phi(a \wedge b) \ge \phi(a ) + \phi(b)$
and let $a_i= \{i \}^\complement$ the complement of  $\{i \}$ in $\{1,2,3,4,5\}$,   then
$$\sum_{r=1}^5 (-1)^{r+1}\sum_{i_1 < i_2 < \cdots < i_r} \phi( a_{i_1}  \wedge \cdots \wedge a_{i_r}) \ge 0$$ proof: we reorganize the alternative sum into the sum of the following components:

*

*$\phi(\{1,2,3,4\}) - \phi(\{1,2,3\}) - \phi(\{1,2,4\}) + \phi(\{1,2\})$

*$\phi(\{1,3,4,5 \}) - \phi(\{1,3,4\}) - \phi(\{1,3,5\}) + \phi(\{1,3\})$

*$\phi(\{2,3,4,5\}) - \phi(\{2,3,4\}) - \phi(\{3,4,5\}) + \phi(\{3,4\})$

*$\phi(\{1,2,4,5 \}) - \phi(\{2,4,5\}) - \phi(\{1,4,5\}) + \phi(\{4,5\})$

*$\phi(\{1,2,3,5\}) - \phi(\{1,2,5\}) - \phi(\{2,3,5\}) + \phi(\{2,5\})$

*$  \phi(\{1,5\}) - \phi(\{1\})$

*$ \phi(\{2,4\}) - \phi(\{2\})$

*$  \phi(\{2,3\}) - \phi(\{3\})$

*$ \phi(\{1,4\}) - \phi(\{4\})$

*$  \phi(\{3,5\}) -  \phi(\{5\})$

*$ \phi(\emptyset)  $
but the first five components are positive by $(2)$, the next five components are positive by $(1)$, and the last is positive by definition $\square$.
Now the answer of the question is yes by observing that the map $\phi$ defined by $$\phi( a_{i_1}  \wedge \cdots \wedge a_{i_r}) =  \dim (V_{i_1}  \cap \cdots \cap V_{i_r})$$ checks $(1)$ and $(2)$. For $(1)$ it is immediate. For $(2)$ we use the following equality and inclusion: $\dim(U+V) = \dim(U) + \dim(V)  - \dim(U \cap V)$ and $(A\cap B ) + (A\cap C) \subseteq A$.
A: Here is proof by hand for $n=4$:
Let $X = \bigcap_{i=1}^n V_i$  then by taking $V_i' = V_i \ominus X$ we get that 
$$\sum_{r=1}^n (-1)^{r+1} \sum_{  \ i_1 < i_2 < \cdots < i_r } \dim(V_{i_1} \cap \cdots \cap V_{i_r}) = \dim(X) + \sum_{r=1}^n (-1)^{r+1} \sum_{  \ i_1 < i_2 < \cdots < i_r } \dim(V'_{i_1} \cap \cdots \cap V'_{i_r})$$ So we can assume $\dim(X) = 0$.   
First we apply four times the inequality:  $ \dim(U +V + W) \le $ $$ \dim U + \dim V + \dim W - \dim (U \cap V) - \dim (U \cap W) - \dim (V \cap W) + \dim(U \cap V \cap W) $$ and we get   $$\alpha  \ge \sum_{  \ i_1 < i_2 < i_3 } \dim(V_{i_1} + V_{i_2} + V_{i_3})-2\sum_i\dim(V_i) + \sum_{  \ i_1 < i_2 } \dim(V_{i_1} \cap V_{i_2} )$$ Next we apply six times the equality: $$\dim(U+V) = \dim U + \dim V - \dim (U \cap V)$$ and we get
$$\alpha  \ge \sum_{  \ i_1 < i_2 < i_3 } \dim(V_{i_1} + V_{i_2} + V_{i_3}) - \sum_{  \ i_1 < i_2 } \dim(V_{i_1} + V_{i_2} ) + \sum_i\dim(V_i)$$ Finally we observe that:
$\dim(V_1+V_2+V_3) \ge \dim(V_1+V_3)$
$\dim(V_1+V_2+V_4) \ge \dim(V_2+V_4)$
$\dim(V_1+V_3+V_4) \ge \dim(V_1+V_4)$
$\dim(V_2+V_3+V_4) \ge \dim(V_2+V_3)$
$\sum_i\dim(V_i) \ge \dim(V_1+V_2) + \dim(V_3+V_4)$   
It follows that $\alpha \ge 0$   $\square$
