Does this arrangement of polygons necessarily contain a hole?

Question

Consider an arrangement of finitely-many open polygons in the plane (not necessarily convex) such that each polygon intersects at least two other non-intersecting polygons.

Is there always a sub-arrangement such that the union of all polygons in the sub-arrangement contains a hole?

In the illustration below, the blue arrangement on the left satisfies the requirements and indeed it contains a hole; the brown arrangement on the right does not satisfy the requirements (each polygon intersects two other polygons, but these other polygons intersect), and it does not contain a hole.

Answer 1

Consider vertical and horizontal stripes, like in $8\times 8$ checkerboard. Each stripe intersects four pair-wise disjoint stripes, but the board has no holes.

Note

The answer above was given to the question in its first revision.

Subsequent edits added explicit requirement for the sets in question to be closed first, and then to be open, which rendered the answer irrelevant.

This looks like a chase, so fixing an answer doesn't make sense.
I restored it to its initial shape and do not care about it any more.

Answer 2

I might have an approach for convex polygons. I am not sure if I understand your question correctly, so let me give an answer in the case of convex polygons for both of my interpretations:

1. assuming that any set has to intersect two other sets which don‘t intersect each other, but it could also intersect pairs that do.

In that case, there does not need to be a hole, take e.g. your blue example and add a set that covers the hole. CiaPan‘s answer also gives another counterexample.

2. assuming that each set intersects at least two other sets and any pair it intersects is non-intersecting.

In that case there has to be a hole: there is no triple of sets with a common intersection, so the nerve complex is 1-dimensional, i.e., a graph. It cannot be a tree (as otherwise some set would only intersect one other set), so it has some cycle. By the nerve theorem, the union of the sets thus has a hole.

I can imagine that a similar argument might work also for non-convex polygons, but I haven‘t thought i through.

Answer 3

I believe the following approach should work for all polygons (I assumed that (1) the polygons are open, (2) existence of holes refers to a non-trivial first homology group and (3) the condition means that all polygons intersecting a fixed polygon are mutually disjoint):

Let $\mathcal{P}=\{P_1,\ldots,P_n\}$ be the polygons. We first show that the union of some subsets of $\mathcal{P}$ has a hole.

By the assumption, re-ordering if necessary, we can assume that for some $(4\le) k\le n$ and for all $1\le i<j\le k$, $P_i\cap P_j\neq\emptyset $ iff $j=i+1$ or $(i,j)=(1,k)$. Using the Mayer-Vietoris sequence (for reduced homology) and the fact that polygons are contractible, we have:

If $P_i\cap P_{i+1}$ is not connected for some $i\in\{1,\ldots,k-2\}$, take the minimum such $i$, then $H_1(P_1\cup \cdots \cup P_{i+1})\simeq\tilde{H}_0(P_i\cap P_{i+1})$ is non-trivial. Otherwise, $H_1(P_1\cup \cdots \cup P_k)\simeq \tilde{H}_0((P_1\cup P_{k-1})\cap P_k)$, which is non-trivial by construction.

Suppose $H_1(\bigcup \mathcal{P})=0$, let $m$ be the smallest integer such that $H_1(P_1\cup\cdots\cup P_m)=0$. Since $(P_1\cup\cdots\cup P_{m-1})\cap P_m$ is a disjoint union of open polygons, it has a trivial first homology. Once again by Mayer-Vietoris, $0=H_1(P_1\cup\cdots\cup P_{m-1})\oplus H_1(P_m)\simeq H_1(P_1\cup\cdots\cup P_{m-1})$, a contradiction.

Edit: A later edit to the question has clarified that my assumption (3) is not what was intended. In any case, I will leave this answer as it is, because the question under assumption (3) is also a very interesting one.