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I am having trouble with the concept of convexity. This is the statement I'm trying to prove.

Let $X$ be normed vector space. If $S \subset X$ is convex and $S^\circ \neq \emptyset$ then $\partial \overline{S} = \partial S$.

Here, $S^\circ$ denotes the interior of $S$, $\overline{S}$ is the closure of $S$, and $\partial S$ is the boundary of $S$.

The definition of convexity I'm given in my textbook is as follows: The set $S \subset X$, where X is vector space, is convex if for all $x,y \in S$ and for all $t \in (0,1)$ it is true that $tx + (1-t)y \in S$.

My approach was to prove both $\partial \overline{S} \subset \partial S$ and $\partial \overline{S} \supset \partial S$. I had no trouble with the first, as I didn't need to use convexity at all. However, I cannot figure out how to prove the other. I've tried it like this:

Let $x$ be in $\partial S$. In order to prove that $x \in \partial \overline{S}$, we must show (by definition of boundary given in my textbook) that for all $\varepsilon>0$ the open ball $B(x,\varepsilon)$ intersects with both set $\overline{S}$ and its complement $X\setminus \overline{S}$.

Yet again I hit a dead end trying to prove the second one, because I do not understand how convexity plays into it. (I have some sort of intuitive understanding of why the initial statement holds true, but I can't quite figure out how prove it.)

Perhaps some other approach is best instead. I would appreciate any help!

aio
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1 Answers1

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I'd like to suggest the following approach:

Since $S$ has non-empty interior, there is a point $x$ and an $\varepsilon>0$ such that $B_\varepsilon(x)\subseteq S$. Given any $y\in S$, one can show that every point $z=ty+(1-t)x$ for $0\le t<1$ is in the interior of $S$. If you'd like to prove this, here is a hint: Show that any $w\in B_{(1-t)\varepsilon}(z)$ is in $S$ by showing that $v:= x+(1-t)^{-1}(w-z)$ is in $S$ and $w$ is on the line connecting $v$ and $y$.

Now given $b\in\partial S$, it cannot lie on a line between any point in int$(S)$ and any point in $S\setminus\{b\}$. Hence the "cone" $O_b = \{tb+(1-t)u \mid u\in\text{int}(S), t>1\}$ is disjoint from $S$. Can you show that

  • $O_b$ is open (and thus disjoint from $\overline S$),
  • every neighborhood of $b$ intersects $O_b$.
Stefan Hamcke
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  • I was able to follow your approach up until the "cone" came to play (thank you for the hint for that proof!). I am rather new to this topic and I don't quite understand the concept of the "cone" and what it will give me if I show those two properties. If you could explain what exactly $O_b$ is and what those properties give me, I'm sure I can figure out the rest of the proof on my own. Thank you! – aio Mar 13 '16 at 13:53
  • @aio: That's not the official name of this construct, I just chose that term because the set looks like a cone (with apex $b$). To get a feeling for this set, I suggest you draw for example the unit square in $\Bbb R^2$ and look at $O_b$ for varying $b\in\partial I^2$. Do you see what $O_{(0,0)}$ looks like? How about $O_{(1/2,0)}$? – Stefan Hamcke Mar 13 '16 at 15:09
  • To finish the proof using $O_b$, note that if $O_b$ is disjoint from $\overline S$ and every neighborhood of $b$ intersects $O_b$, then this just means that $b$ is in the boundary of $\overline S$. @aio – Stefan Hamcke Mar 13 '16 at 15:12
  • I'm sorry, I feel really dumb, but I still don't understand what $O_b$ looks like (or what it is). And why does the fact that $O_b$ is open mean that $O_b$ is disjoint from $\overline S$? For the end of the proof, I see that, if I show $O_b$ is disjoint from $\overline S$, then $O_b \subset X \setminus \overline S$. Then, if I show that every neighborhood of $b$ intersects $O_b$, I will have $b \in \partial \overline S$. – aio Mar 13 '16 at 16:35
  • @aio: Since $O_b$ is disjoint from $S$ and open, any point $x\in O_b$ has a neighborhood not intersecting $S$, hence $x$ is not in the closure of $S$. – Stefan Hamcke Mar 13 '16 at 16:41
  • @aio: To get a feeling for $O_b$, pick some $u$ in the interior of $S$ and draw a ray $\ell$ through $u$ and $b$. The ray is divided into three parts: The line between $u$ and $b$, the ray starting at $u$, and the ray starting at $b$. The later ray minus $b$ is then in $O_b$, and $O_b$ consists of all these open rays. For the unit square and $b=(0,0)$ this is the open quadrant where $x$ and $y$ are negative, for $b=(1/2,0)$ it is the lower open half-plane where $y$ is negative. – Stefan Hamcke Mar 13 '16 at 16:49
  • @aio: You could think of $S$ as a hollow box, $b$ is a hole in the box, and $O_b$ is the space from where you can look through $b$ and observe the inside of the box. I hope this makes sense ... – Stefan Hamcke Mar 13 '16 at 17:13
  • Yes, thank you so much for your explanation and patience, it makes much more sense now! All I have to do now is chew through the proof of those two properties. I'm running into some difficulty there as well (though they are quite obvious intuitively), but I'm sure I'll manage it eventually. – aio Mar 13 '16 at 17:45
  • @aio: You are welcome. You should try to prove the properties by yourself. But if you really struggle, I can provide some hints. The proofs are actually really short, it's just a matter of having the right idea :-) – Stefan Hamcke Mar 13 '16 at 17:54
  • Hints would be much appreciated after all, can't quite figure it out. – aio Mar 13 '16 at 18:24
  • @aio: For openness of $O_b$: Since $S°$ is open, so is $tb+(1-t)S°$ when $1-t\ne0$ and thus also $O_b=\bigcup_{t>1}tb+(1-t)S°$ – Stefan Hamcke Mar 13 '16 at 18:53
  • @aio: For the other property: Given $ε>0$ and $u\in S°$, show that $$ \left( 1+\fracε{||b-u||} \right)b - \fracε{||b-u||}u $$ is in $O_b\cap B_{2ε}(b)$. – Stefan Hamcke Mar 13 '16 at 18:58