Let $n=3.$

Take $X=(0,1)^n.$ Fix points $p,q$ s.t. $\text{dist}_n(p,q)=\sqrt{n}.$ Construct a smooth regular foliation of $X$ with $(n-1)-$dim. leaves which are topologically $(0,\sqrt{n})\times S^{n-2} $ accumulating to $p,q.$

In this 2021 paper by Inaba and Masuda Foliations on the open 3-ball by complete surfaces they describe classes of surfaces that foliate $X=(0,1)^3$ without any singularities. One class they describe is $S=\Bbb R^2 - (C \times \lbrace{0}\rbrace)$ where $C$ is the Cantor set. Another class is $S=\Bbb R^2-(\Bbb Z \times \{0\}).$

My formulation reduces to the class $S=\Bbb R^2- (\text{1 point})$ which the authors do not consider. It's curious that Inaba and Masuda have found weirder examples than mine which *do* foliate $X=(0,1)^3$ without any singularities. This suggests that it might be possible to achieve such a foliation.

Note that the leaves in $X$ are all diffeomorphic to open annuli, and the open annuli are diffeomorphic to once punctured planes. This is why the problem is equivalent to asking about a foliation by once punctured planes.

How do you construct a foliation of $X$ with the class $S=\Bbb R^2- (\text{1 point})$ with no singularities?

openinterval, so (0, 1)^3 is anopenunit 3-cube. My question was:What set or sets are p and q members of?(Not: "What are examples of p and q?") $\endgroup$distinctvertices on the $n$-cube (all else equal), then the problem still is equivalent to a foliation of $X$ by once punctured planes (in dimension $n=3$) and corresponding higher dimensional equivalents. So the formulation I gave is simply another way of stating the problem. $\endgroup$openn-cube and [0, 1]^n denotes theclosedn-cube. The open version (0, 1)^n does not contain any of its boundary points. $\endgroup$1more comment