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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.

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  • $\begingroup$ @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.) $\endgroup$
    – Gabriel
    Aug 18, 2019 at 16:41

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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.

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    $\begingroup$ But $V=I \times A$ already, for $I=(0,+\infty)$ and $A=S^{k-1}\backslash\{p\}$ $\endgroup$ Aug 18, 2019 at 17:38
  • $\begingroup$ @ThiagoLandim No, we want to have an infinite disjoint union. If you also allow finite unions, then take $V = ((0,\infty) \times S^{k-1}) \setminus \{ p \}$. $\endgroup$
    – Paul Frost
    Aug 18, 2019 at 17:40
  • $\begingroup$ 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. $\endgroup$ Aug 18, 2019 at 18:25
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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 $H^2(U;\mathbb Z) \approx \mathbb Z^2$.

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  • $\begingroup$ You prove much more than the OP expected, but very nice! (+1). $\endgroup$
    – Paul Frost
    Aug 18, 2019 at 17:37

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