How to prove that for a nonempty convex subset $S \subset X$ ($X$ is normed vector space) it is true that $\partial \overline{S} = \partial S$?

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!

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 $O_b$.
• 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