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. 
 A: 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.
A: 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 $H^2(U;\mathbb Z) \approx \mathbb Z^2$.
