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For proving Jordan separation theorem in differential manifold theory, one step involves proving the following: Let $z\in{\mathbb{R}^n}\setminus X$, where $X$ is connected, closed manifold of dimensin $n-1.$ suppose that $x\in{X}$ and $U$ is an arbitrary neighborhood of $x$. Then we have to show that there is a point of $U$ that may be joined to z by a curve not intersecting X.\

If we take $S$ to be the set of all those points of $X$ with this property and prove it open and cloed, then using connectedness of $X$ it will be whole of $X$. I have showed it to be closed. Can anyone help me with the openness part of this.

user1123
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  • What exactly do you want to prove? – Peter Franek May 22 '14 at 19:11
  • @PeterFranek i want to prove that any point of in a neighborhood of any point x of $X$ can be joined to $z$ without intersecting $X$. And this fact is used to further prove the theorem – user1123 May 22 '14 at 19:13
  • @PeterFranek please tell me if question is unclear. – user1123 May 22 '14 at 19:23
  • I removed the downvote, sorry. Didn't know that this is called "Jordan brouwer separation theorem". Will think about some simple proof. – Peter Franek May 22 '14 at 19:57
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    For the openness, I think you can just take a neighborhood $U(x)$ of $x$ homeomorphic to $\mathbb{R}^n$ s.t. $U(x)\cap X$ corresponds to $\mathbb{R}^{n-1}\times{0}$ and assume wlog that points in the "upper half-space" of $U(x)$ can be connected to $z$ in $\mathbb{R}^n\setminus X$ (if one can be connected, then any). Then, if $y\in\mathbb{R}^{n-1}$ is any other point in $U(x)\cap X\simeq \mathbb{R}^{n-1}$, then any neighborhood of $y$ contains a point in the upper halfspace and can be joint to $z$ as well. – Peter Franek May 22 '14 at 20:38

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