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Assume that we have a codimension one foliation of $\mathbb{R}^{n}-\{0\}$ with compact leaves.

Is it true to say that the foliation is stable at origin:That is: for every neighborhood $V$ of $0$, there is a neighborhood $W\subset V$ containing $0$ such that the saturation of $W$ is contained in $V$?

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Yes. Let $\mathcal{F}$ be the foliation (of arbitrary dimension in $\{1,\ldots,n-1\}$), which I assume regular and transversely continuous. It is sufficient to consider the case of a ball $V$ of radius $r>0$. I claim that the set $A:=\mathrm{Sat_\mathcal{F}}(\partial{V})$ is compact, therefore $V\setminus A$ is a non-empty open set $W$, invariant by $\mathcal{F}$, giving the sought neighbourhood.

The fact that $A$ is compact follows from an elementary flow-box argument. For every $p\in \partial V$ there exists a small compact neighbourhood $V_p\ni p$ for which $\mathrm{Sat}_\mathcal{F}(V_p)$ is compact (the foliation is locally a $C^0$-fibration). By compactness of the sphere $\partial V$, the set $A$ is the union of finitely many such $\mathrm{Sat}_\mathcal{F}(V_p)$, thus compact.

Remark: The argument works as soon as one removes from $\mathbb{R}^n$ any compact set $X$ and consider neighbourhoods $V$ of $X$. This hypothesis is somehow optimal: as the Hopf fibration shows, this is false for $n=3$ and $X$ a line.

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  • $\begingroup$ Are you sure that the argument works for one dimensional foliation? Can not we imagine a non vanishing vector field on R^n with a dense orbit $\gamma$. Now if we remove a point $p\notin \gamma$ then we have a foliation of R^n-p which is not satable at p. Am I mistaken? $\endgroup$
    – Ali Taghavi
    Commented Feb 19, 2016 at 19:55
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    $\begingroup$ But a dense orbit is not compact, is it ? Anyway, here the argument is that $W$ cannot cross $\partial{V}$. $\endgroup$
    – Loïc Teyssier
    Commented Feb 19, 2016 at 21:05
  • $\begingroup$ I was wrong. Another question: could you please more explain on "For every $p\in \partial V$ there exists a small compact neighbourhood $V_p\ni p$ for which $\mathrm{Sat}_\mathcal{F}(V_p)$ is compact " What is the reason?Sorry if my question is elementary. $\endgroup$
    – Ali Taghavi
    Commented Feb 20, 2016 at 12:04
  • $\begingroup$ Well, since the vector field is regular it admits local flow-boxes. Since a leaf is compact, it suffices to consider finitely many such flow-boxes to cover the leaf $L_p$ passing through $p$. One can then choose the flow-boxes in such a way that the foliation is a locally trivial fibration over $U\to L_p$. Another way of seeing this is to take a compact subset $\Sigma\subset\partial{V}$ and consider the compact product $\Sigma\times L_p$ and prove (up to taking a finite covering) that it is homeomorphic to $U$ (because the foliation is transversely continuous). $\endgroup$
    – Loïc Teyssier
    Commented Feb 21, 2016 at 10:57
  • $\begingroup$ Thank you very much for your answer.. Your answer is interesting. It remind me the local or global fibration discussion in the following, which essentially use the assumption compactness of all leaves. mathoverflow.net/questions/150735/… $\endgroup$
    – Ali Taghavi
    Commented Feb 22, 2016 at 17:16

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