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I need to show that if in a normed space X, a set $A \subset X$ is convex and its interior is non-empty, then it has the same boundary as its closure: $\partial \overline A = \partial A$. I would like to formulate a proof that uses only the definition of convexity: $\forall x,y \in A, \lambda \in (0,1): \lambda x + (1-\lambda)y \in A$.

I'm able to show that $A \subset X \implies \partial \overline A \subset \partial A$, but I can't figure out how to use the definition of convexity to show that $\partial \overline A = \partial A$.

Thank you very much :)

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  • $\begingroup$ The assumption that $A$ has nonempty interior is essential here. The idea is that if $x$ is in the interior of $\overline{A}$, you can draw all the line segments from points of $A$ near $x$ to points in the interior of $A$, and these will cover an entire ball around $x$, so $x$ has to be in the interior of $A$. $\endgroup$ – Eric Wofsey Mar 13 '16 at 19:19
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To get $\partial A\subseteq\partial\overline A$, it is enough to show that the interior of $\overline A$ is contained in the interior of $A$.

First, note that it is in fact enough to show that the interior of $\overline A$ is contained in $A$. Indeed, assume this has been done, and take any point $a$ in the interior of $\overline A$. Then there is an open ball $B$ centered at $a$ and contained in $\overline A$. Since $B$ is open, it is in fact contained in the interior of $\overline A$, and hence in $A$ by assumption; and this shows that $a$ is an interior point of $A$.

Now let us show that the interior of $\overline A$ is indeed contained in $A$. Let $a$ be any point in the interior of $\overline A$, and choose an open ball $B$ centered at $a$ such that $B\subseteq \overline A$. Let also $x_0$ be an interior point of $A$, and choose an open ball $B_0$ centered at $x_0$ such that $B_0\subseteq A$. Consider a point $x'_0\in B$ such that $a\in [x_0,x_0')$ and very close to $a$, close enough to ensure that the homothety $h$ with center $a$ such that $h(x_0)=x'_0$ maps the ball $B_0$ onto a ball $B'_0$ contained in $B$. Then $B'_0\subseteq\overline A$, and since $B'_0$ is open, it follows that one can find a point $x'\in B'_0$ such that $x'\in A$. Then $x'=h(x)$ for some point $x\in B_0$. By the definition of $h$, we then have $a\in [x,x')$, and hence $a\in A$ since both $x$ and $x'$ are in $A$.

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