# Is any open set of $(0,\infty)\times S^{k-1}$ a countable disjoint union of products $I\times A$, where $I$ is an interval and $A$ is open?

Let $$V$$ be an open set in $$(0,\infty)\times S^{k-1}$$. I want to prove that $$V$$ can be written as a countable disjoint union of the form $$V=\coprod_{n=1}^\infty I_n\times A_n,$$ where each $$I_n$$ is an open interval and each $$A_n$$ is open in $$S^{k-1}$$.

I think this is true since we can write any open set in $$(0,\infty)$$ as a countable disjoint union of intervals and $$(0,\infty)\times S^{k-1}$$ is second countable. However I was not able to write a proof.

Edit: By second-countability, we can write $$V=\bigcup_{n=1}^\infty U_n\times A_n,$$ where $$U_n$$ is open in $$(0,\infty)$$ and $$A_n$$ is open in $$S^{k-1}$$. Now. for each $$n$$ we can write $$U_n=\coprod_{k=1}^\infty I_k^n,$$ where each $$I_k^n$$ is an interval. This implies that $$V=\bigcup_{n=1}^\infty \coprod_{k=1}^\infty I_k^n\times A_n.$$ This is clearly a countable union of the desired form but I'm not sure it is disjoint.

• @Dzoooks this is not clear to me. By the definition of product topology we can write $V$ as a union of $U_\alpha\times A_\alpha$, where $U_\alpha$ and $A_\alpha$ are open. Also, for every $\alpha$, we can write $U_\alpha$ as a countable disjoint union of intervals. But I don't know how we can continue from that. (For sure, we can make the union countable. I'm just not sure about it being disjoint.) Commented Aug 18, 2019 at 16:41

In general no. Pick a point $$p \in S^{k-1}$$ and let $$V = (0,\infty) \times (S^{k-1} \setminus \{ p \})$$. We have $$V \approx (0,\infty) \times \mathbb R^{k-1}$$ which is pathwise connected, hence connected.

Assume that there is a solution. The $$A_n$$ must be contained in $$S^{k-1} \setminus \{ p \}$$ and are thus open in $$S^{k-1} \setminus \{ p \}$$. Hence each $$I_n \times A_n$$ is open in $$V$$. You thus get $$V = I_1 \times A_1 \cup \bigcup_{n=2}^\infty I_n \times A_n$$ which is a partition of $$V$$ into two nonempty disjoint open sets. This means that $$V$$ is not connected, a contradiction.

• But $V=I \times A$ already, for $I=(0,+\infty)$ and $A=S^{k-1}\backslash\{p\}$ Commented Aug 18, 2019 at 17:38
• @ThiagoLandim No, we want to have an infinite disjoint union. If you also allow finite unions, then take $V = ((0,\infty) \times S^{k-1}) \setminus \{ p \}$. Commented Aug 18, 2019 at 17:40
• The empty set is open, so you may take $I_n =A_n=\varnothing$ for $n \geq 2$. But you are right: I believe it is simpler to prove the falsehood for $\mathbb{R}^k$ and use the natural embedding of $\mathbb{R}^{k-1}$ in $S^{k-1}$ to conclude for this case. Commented Aug 18, 2019 at 18:25

I'll describe a counterexample.

But first, arguing by contradiction, if what you were saying were true, then the following consequences would ensue for every connected open subset $$U \subset (0,\infty) \times S^{k-1}$$:

• If $$U$$ is connected then $$U$$ is homeomorphic to $$I \times A$$ for some interval $$(0,1) \times A$$ where $$A \subset S^{k-1}$$ is open (because all the terms of your disjoint union would be open sets, so if there were more than two terms then the subset would be disconnected).
• $$U$$ is homotopy equivalent to $$A$$ (because the projection map $$I \times A \mapsto A$$ is a homotopy equivalence).
• $$H_{k-1}(U;\mathbb Z)$$ is either trivial or infinite cyclic (because if $$A \subset S^{k-1}$$ is proper then its $$H^{k-1}(A;\mathbb Z)$$ is trivial; otherwise $$H^{k-1}(A;\mathbb Z)$$ is infinite cyclic.

So, all I have to do is to describe for you a connected open subset $$U$$ for which $$H^{k-1}(U;\mathbb Z)$$ is ismorphic to $$\mathbb Z^2$$. Here's an example in dimension $$k=3$$, and it easily generalized to higher dimensions: $$U = ((0,1) \times S^2) \cup (0,3) \times S^2_+ \cup (2,0) \times S^2$$ where $$S^2_+$$ denotes the open upper hemisphere. A simply Mayer-Vietoris calculation proves that $$H^2(U;\mathbb Z) \approx \mathbb Z^2$$.

• You prove much more than the OP expected, but very nice! (+1). Commented Aug 18, 2019 at 17:37