Cover the $n$-disc with $n+1$ open sets $D^n = U_0 \cup \cdots \cup U_n$. Suppose that $U_0 \cap \cdots \cap U_n = \emptyset$.

Question: Then is some $n$-fold intersection empty? That is, after reordering do we have $U_1 \cap \cdots \cap U_n = \emptyset$?

I believe the answer is yes.

Notes:

• When $n = 1$, this follows from the fact that the disk is connected.

• In general, we can look at the Cech nerve of the cover. We have that the geometric realization is contractible: $|\amalg_{i_0 < \cdots < i_\bullet} U_{i_0} \cap \cdots \cap U_{i_\bullet}| \equiv |D^n| \equiv \ast$. Since the $(n+1)$-fold intersection is empty, there is a map from the geometric realization the geometric realization of the $(n+1)$-cube with top and bottom removed (i.e. what you’d get from the Cech nerve if all the intersections are a point except for the $(n+1)$-fold one which is empty), which has homotopy type $S^{n}$. I suspect this map is always essential if all of the $k$-fold intersections $U_{i_1} \cap \cdots \cap U_{i_k}$ are nonempty for $k \leq n$, which would be enough to prove an affirmative answer to my question.

• When $n = 2$, I believe that the intersections $U_{i_0} \cap \cdots \cap U_{i_k}$ must have the homotopy types of disjoint unions of wedges of circles (because they are open subsets of $D^2$). Then I think I can convince myself that the Cech nerve is homotopy equivalent to something you can build up one cell at a time, and that it must have a wedge summand of $S^1$ (more precisely, the map from the previous bullet splits). But this is a bit messy, and will get messier as the dimension increases.