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Let $X$ be a normed space and $N$ a convex subset of $X$ (also $N^\circ \neq \emptyset$). I am trying to show that $\partial \bar N = \partial N$. I found the proof that $\partial \bar N \subset \partial N$ from The topological boundary of the closure of a subset is contained in the boundary of the set (the last comment), but I can't figure out the proof for $\partial N \subset \partial \bar N$.

This is what I've figured out: Let $x \in \partial N$. We want to show that if $U$ is an open neighbourhood of $x$ then $U \cap \bar N \neq \emptyset$ and $U \cap (X \backslash \bar N) \neq \emptyset$. Since $x \in \partial N$ then $U \cap N \neq \emptyset$ and $U \cap (X \backslash N) \neq \emptyset$. Since $N \subset \bar N$ and $x \in \partial N$ then $U \cap \bar N \neq \emptyset$. How do I show that $U \cap (X \backslash \bar N) \neq \emptyset$? I've also noticed that I haven't used the fact that $N$ is a convex set.

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You do need to use the fact that it's convex with nonempty interior. For example, in $X = \mathbb R^2$ the complement of a line gives a counterexample: its boundary is the line, but the boundary of its closure is empty.

Hint: by the Hahn-Banach separation theorem, there is a hyperplane containing $x$ such that ...

EDIT: OK, here's a hint for a proof without the separation theorem. Suppose $x \in \partial N$. Show that $2x - N^\circ = \{2x-y: y \in N^\circ\}$ is open and disjoint from $N$, and contains points arbitrarily close to $x$.

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  • $\begingroup$ I found several solutions using the separation theorem, but I'd like a proof that doesn't involve said theorem or hyperplanes. Is there any way to solve this only using basic topology properties (boundary, closure properties, convex set definition, open set definition etc)? $\endgroup$ – Divine Mar 11 '16 at 22:57
  • $\begingroup$ Let $x \in \partial N$. Then for every $r > 0$ $B(x,r) \cap N \neq \emptyset$ and $B(x,r) \cap (X \backslash N) \neq \emptyset$. Lets look at the set $A := \{ 2x - y : y \in N^\circ \}$. Let $z := 2x - y$. Since $y \in N^\circ$ then $\exists \rho > 0 : B(y, \rho) \subset C$. Both $B(x,r)$ and $B(y, \rho)$ are open. Now I don't know what direction to go in because I'm not sure how all these balls intersect (or if they even do). $\endgroup$ – Divine Mar 12 '16 at 16:40
  • $\begingroup$ If $z = 2 x - y$ and $B(y,\rho) \subset N$, then $B(z,\rho) \cap N = \emptyset$, since if $w \in B(z,\rho) \cap N$, $(w + B(y,\rho))/2$ would be a neighbourhood of $x$ contained in $N$. $\endgroup$ – Robert Israel Mar 13 '16 at 7:24
  • $\begingroup$ So if there was a $w \in B(z, \rho) \cap N$ then we would get a contradiction with the fact that $x \in \partial N$ because $(w + B(y, \rho))/2$ is a neighbourhood of $x$ contained in $N$ (which means that $x \in N^\circ$) I'm not sure how to show that $(w + B(y, \rho))/2$ is a neighbourhood of $x$ because I don't understand the notation $(w + B(y, \rho))/2$. $\Vert z-x \Vert = \Vert x - y \Vert \leq ?$ We can conclude that $A \cap C = \emptyset$. I don't know how to show that $A$ is open and I'm not sure how I would use those facts (& the convex set definition) to get $x \in \partial \bar N$ $\endgroup$ – Divine Mar 13 '16 at 16:14
  • $\begingroup$ I've been trying my best to figure this out but I'm finding it quite difficult. I'm pretty sure that I can use the fact that $N^\circ$ is open to show that $A$ is open but I can't figure out how to do that. Does $(w + B(y, \rho))/2$ mean the ball $1/2 B(w+y, \rho)$? So if $A$ is open and disjoint from $N$ then $A \subset X \backslash N$ (and $A \subset X \backslash \bar N$?). Does showing that every neighborhood of $x$ intersects $A$ give me that $x \in \partial \bar N$? If so, how could I show that to be true? Also I'm still now sure where I will have to use the fact that N is convex. $\endgroup$ – Divine Mar 13 '16 at 22:33

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