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In this question, one of the answer (by Dimitris) uses the following lemma.

Lemma. Let $A\subset \mathbb{R}^N$ be a convex set. Suppose $\text{Int}A \ne \emptyset$. If $x \in \text{Int} A$ and $y \in \text{Cl}A$, then $[x, y) \subset \text{Int}A$, where $$ [x,y) := \{y+\lambda(x-y)\mid\lambda\in(0,1]\}. $$

(I changed the statement in the case of $\mathbb{R}^N$ since it suffices for my purpose.)

I am having trouble in proving this lemma. Here is my attempt.

Pick any $x \in \text{Int} A$ and $y \in \text{Cl}A$. If $y \in \text{Int}A$, then the result follows because $A$ is convex and thus $\text{Int}A$ is as well. Suppose $y \in \text{Bdry}A:=\text{Cl}A \setminus \text{Int}A$. Pick any $z \in [x,y)$. Then, there exists $\lambda\in(0,1]$ such that $z = y + \lambda(x-y)$. If $z \notin \text{Int}A$, then $z\in \text{Bdry}A$ since $z \in \text{Cl}A$ by the convexity of $A$ and $\text{Cl} A$.

Although it is intuitively clear that $z,y\in\text{Bdry}A$ is a contradiction, how can I prove it?

keepfrog
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2 Answers2

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Case 1: assume also $y\in A$. Since $x \in \text{Int} A$, there exists $\rho>0$ such that $B_\rho(x)\subset \text{Int} A$. Let $z = (1-\lambda) y + \lambda x$, $\lambda\in (0,1)$. Then $B_{\lambda\rho}(z) = z + B_{\lambda\rho}(0) = (1-\lambda)y + \lambda B_\rho(x) \subset A$, so that $z$ is an interior point of $A$.

Case 2: assume $y\in \text{Cl} A$. The ball $B_{\rho\lambda/(1-\lambda)}(y)$ contains a point $a\in A$; in particular, $a = \frac{\lambda}{1-\lambda} u + y$, for some $u\in B_\rho$. Hence, $z = (1-\lambda) a + \lambda(x-u)\in A$. By case 1, we conclude that $z\in \text{Int} A$.

Rigel
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  • Why can you say that $(1-\lambda)y + \lambda B_{\rho} (x) \subset A$? If $y \in A$, then it is clear from the convexity of $A$, but what if $y \in \text{Bdry} A$? – keepfrog Jul 19 '20 at 10:04
  • Right, see the edited answer. – Rigel Jul 19 '20 at 10:16
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I'll assume $N\geq 2$ (the case $N=1$ is a lot simpler). Let $r>0$ be such that the closed $r$ ball around $x$ is in $A$. By rotating and shifting, I may assume that $x$ is the origin, and that $y$ lies on the second axis.

Let $z_n$ be a sequence of points in $A$ approaching $y$. We know $p_1:=(r,0,...0)$ and $p_2:=(-r,0,...0)$ are both in $A$, so, by convexity, the triangle $\triangle_n$ with vertices $z_n,p_1,p_2$ lies entirely inside $A$.

Now, let $p$ be a point on $[x,y)$. I think it shouldn't be hard to show that for large enough $n$, the point $p$ must be in the interior of $\triangle_n$, which means it must be in the interior of $A$.

Cronus
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  • It seems that if $x$ is the origin and $y$ lies in the first axis, then $y,p_1,p_2$ does not form triangle in $\mathbb{R}^2$ (they are all on the first axis). Moreover, "some details" you mentioned might be essentially the same as my original question. – keepfrog Jul 19 '20 at 10:08
  • @user789100 Sorry, sorry, you're right. I'll edit the question. – Cronus Jul 19 '20 at 10:11
  • I edited the question - is it clear now? – Cronus Jul 19 '20 at 10:13
  • Suppose $A$ is open and $y \in \text{Bdry} A$. Then, the vertex $y$ is not contained in $A$. In that case, how can you say that $[x,y)$ is in the interior of $A$? I am guessing that proving this point is essentially the same as completing my original attempt... – keepfrog Jul 19 '20 at 10:32
  • @user789100 I'm really sorry, I misread your question. I should have read more carefully. I tried to correct my answer, but there is still a little gap. If it's completely useless to you I can delete my answer. – Cronus Jul 19 '20 at 10:40
  • Thanks for trying to answer the question. The edited version seems make sense to me. If $y\in\text{Cl}A$, by taking $z^m = B_{1/m} (y) \cap A$ we can obtain the mentioned sequence, and by observing $ | y - p | = \lambda | x - y | >0$ I think we can obtain the desired result. – keepfrog Jul 19 '20 at 10:55
  • @user789100 Yeah, I think something like that should work. – Cronus Jul 19 '20 at 12:38