Question 0. If a connected, locally path-connected topology X cannot be covered by two open sets U and V, each connected, so that the intersection U ∩ V is disconnected, must X be simply connected?
I was toying around with different ways to detect holes in a topological space that doesn't assume the given topology is "nice".
For example, consider the long line L = [0..1) × ω₁, but coarsen it in the following way: if a set U contains the point (0,0), then it must contain a non-empty final segment ((r,α)..∞) of L for it to be open. Call the resulting topology the long circle, LS¹.
None of the usual ways of detecting holes in topology (that I know of) can detect a hole in LS¹. The long circle also isn't a particularly nice space. It is path-connected but, unlike the long line, it is not locally path connected. For example, the point (0,0) in LS¹ has neighbourhood LS¹ \ {(0,1)}, any subneighbourhood of that neighbourhood must contain a point (r,α) > (0,1) and there is no path from (0,0) to such point (r,α). However, intuitively, it is just a (maybe weird) circle, and we expect circles to have holes.
I noticed we can cover the circle S¹ with two connected open sets U and V with disconnected intersection U ∩ V. The same is true for the long circle LS¹: we can choose any two distinct points x and y on the long circle, then U := LS¹ \ {x} and V := LS¹ \ {y} are connected open sets covering LS¹ and their intersection U ∩ V is disconnected. This led to the following definition
Definition 1. Let X be a topological space. We say X is 1-connected iff X is connected (has one connected component) and, for any two connected open sets U and V that cover X, U ∩ V is connected as well.
Not the main question, but:
Question 2. Does definition 1 have a (standard) name in the literature?
I wanted to know whether this definition agrees with the usual definition of simply connected (path-connected and every loop on a point is path-homotopic to the constant loop) for nice enough spaces. I managed to prove one direction modulo the following technical lemma.
Lemma 3.\* Let X = [0..1] × [0..1] be the square and ∂X its boundary. Let U and V be open sets covering X. Let x and y be points on the boundary of X. Assume that x and y are in the same connected component of U ∩ ∂X and in the same connected component of V ∩ ∂X. Then, x and y are in the same connected component of U ∩ V.
I have tried to prove this lemma but didn't get very far. I don't think the intersection with ∂X in the assumptions is necessary.
Theorem 4.\* Let X be a simply connected, locally path-connected topology. Then, X is 1-connected.
Proof. Assume, towards contradiction, X is not 1-connected. Let U and V be connected open sets covering X with disconnected intersection. Let x,y ∈ U ∩ V be points not in the same connected component of the intersection. As X is locally path-connected, so are U and V, and connected locally path-connected sets are path-connected. So, there is a path p: x ↝ y in U and a path q: y ↝ x in V. Consider the concatenation w := p;q: x ↝ x, which is a loop on x. As X is simply connected, there is a path-homotopy H: w ⇒ 1ₓ. I.e. H: [0..1] × [0..1] → X is so that H(0,t) = w(t), H(1,t) = 1ₓ(t) = x and H(s,0) = H(s,1) = x. Consider the open sets H⁻¹U, H⁻¹V and H⁻¹(U ∩ V) = H⁻¹U ∩ H⁻¹V in [0..1] × [0..1]. As H is constant x on the top, right and bottom boundary and x ∈ U ∩ V, we have that H⁻¹U and H⁻¹V contain the top, right and bottom boundary of the square. Further, H on {0} × [0..½] (the bottom half of the left boundary) agrees with p, so lies in H⁻¹U, and similarly {0} × [½..1] lies in H⁻¹V. So, the points (0,½) and (1,½) of [0..1] × [0..1] lie in the same connected component of H⁻¹U ∩ ∂[0..1]² and lie in the same connected component of H⁻¹V ∩ ∂[0..1]². Therefore, by lemma 3, they lie in the same connected component C of H⁻¹(U ∩ V). The image of a connected set is connected, so H[C] ⊆ U ∩ V is connected. However, H(0,½) = y ∈ H[C] and H(1,½) = x ∈ H[C], contradicting that x and y lie in different connected components of U ∩ V. ∎
The asterisks are to signify there is a gap in the proof. Of course, this leads to the following question:
Question 5. How to prove lemma 3? Is it true?
The assumption that X is locally path-connected in the theorem above cannot be weakened to just path-connected as the long circle LS¹ shows.
Question 6. How does the answer to question 0 change if we change the assumption that X is locally path-connected, e.g. to path-connected or to metrizable?
Thanks in advance for your help.