TITLE: A version of Lusternik–Schnirelmann category for good open covers
QUESTION [12 upvotes]: Recall that the Lusternik–Schnirelmann category (or LS-category) of a space is the integer $n$ such that there is an open cover by $n+1$ open sets which have nullhomotopic inclusions, and no such cover by fewer open sets. This is quite hard to compute, but for instance, spheres clearly have LS-category equal to 1. 
However, sometimes one is interested not just in open covers by (relatively) contractible open sets, but by covers where all finite intersections of such opens are also (relatively) contractible. This is especially true when dealing with higher stacks, if one is interested in efficient cofibrant replacements.
Thus, one might want to define something intermediate, like level-$n$ LS-category of a space, which is one more than the minimum number of open sets one needs to cover said space, such that intersections of at most $n$ opens is (relatively) contractible. Then a finite good open cover gives an upper bound on the level-$n$ LS-category for all $n$. There is also the analogue for good open covers, namely where all finite intersections are (relatively) contractible.
For instance, for spheres one can take a homeomorphism with the boundary of a topological simplex, then take the open cover by puffing up a face of the simplex by some small amount (this is the analogue of covering a circle by three arcs). Thus the level-$n$ LS-category of the $k$-sphere is bounded above by $k+1$.
Has something like this been considered in the literature before?
My interest stems from considering this problem for Lie groups, in particular matrix Lie groups $O(n,\mathbb{K})$, where for certain cases we have explicit covers of the minimum size. Even something like an upper bound would be good.

REPLY [3 votes]: In some cases the nerve theorem might help (in the strong case without "relatively"; I also assume that some intersections in the question might be empty instead of contractible, hoping that this the intended definition of a good cover). Apropriate version can be found, for example in [A. Hatcher. Algebraic Topology. Cambridge University Press, Cambridge,
2001., Chapter 4.G]. In case of good covers where all intersections are contractible this yields that $X$ is homotopy equivalent to the nerve of the cover. Thus the number of sets in the cover is bounded from below by the number of vertices of the smallest simplicial complex homotopy equivalent to $X$. In case that the contractibility is  assumed only for intersections up to $n$th level, then maybe some more refine versions of the Nerve theorem such as [G. Kalai and R. Meshulam. A topological colorful Helly theorem.
Adv. Math., 191(2):305–311, 2005., Theorem 3.2] can be used. (In particular, the nerves of good covers have been studied intensively in conections with Helly-type theorems.)
For example, the nerve theorem is already sufficient to get the exact bounds for spheres as described in the statement of the question via the following approach. One can get that $n+1$ sets are sufficient to cover the $(n-1)$-sphere as explained in the statement of the question. However, if we put the assumption on contractibility of intersections only up to level $n$, we get that $n+1$ sets are sufficient to cover arbitrary $k$-sphere with $k \geq n-1$ by considering $(k-n+1)$ suspensions of the example with $k=n-1$ (one has to be little careful to keep the sets open). Altogether we get that $Cat^n(S^k) \leq \min(k+1,n)$, where $Cat^n(X)$ is the minimum j such that there is a cover of $X$ by open sets $U_0,…,U_j$ which are contractible, borrowing the notation analogous to the notation of the answer of Mark Grant.
Using the Nerve theorem, one can deduce that 
$Cat^n(S^k) \geq \min(k + 1,n)$: Indeed, for contradiction, let us assume that $Cat^n(S^k) = j$ where $j \leq k$ and $j < n$. Let $\mathcal{U} = \{U_0, \dots, U_j\}$ be the corresponding cover of $S^k$. Since $j  < n$ we get that all intersections are contractible. Using the Nerve theorem, $S^k$ is homotopy equivalent to the nerve of $\mathcal{U}$. However, the simplicial complex with the smallest number of vertices homotopy equivalent to $S^k$ is the boundary of $k+1$-simplex, which has $k+2$ vertices, contradicting $j \leq k$.