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How can one prove that $ \text{cl}(\text{int}(A)) = \text{cl}(A)$, where $ A \subseteq \mathbb{R}$ is convex?

I know that if $A$ is convex, $\text{int(A)}$ and $\text{cl(A)}$ are convex too.

I need help in $ \text{cl}(A) \subseteq \text{cl}(\text{int}(A))$ part only.

Ashot
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1 Answers1

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I assume that the space is $\mathbb{R}^n$ with standard topology.

Your claim is false as it is, e.g. a single point is a convex set, but it's interior is empty, so $\mathrm{cl}(\{p\}) = \{p\} \neq \varnothing = \mathrm{cl}(\mathrm{int}(\{p\}))$. However, under the assumption that the interior is non-empty, that is $\mathrm{int}(A) \neq \varnothing$, your claim is true.

Let $x \in \mathrm{cl}(A)$ and let $y \in \mathrm{int}(A)$ be any point from the interior of $A$. We know that the interior $\mathrm{int}(A)$ is open, so there exists a ball $B(y,\delta) \subset \mathrm{int}(A)$. Moreover $A$ is convex, hence for all $z \in B(y,\delta)$ we have $[xz] \subset A$. This in turn implies that $(xy] \subset \mathrm{int}(A)$, and finally $x \in \mathrm{cl}(\mathrm{int}(A))$.

The above implies $\mathrm{cl}(A) \subset \mathrm{cl}(\mathrm{int}(A))$, of course $\mathrm{cl}(\bullet)$ is monotone with respecto to $\subset$, therefore $\mathrm{cl}(A) = \mathrm{cl}(\mathrm{int}(A))$.

Have fun ;-)

dtldarek
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    Note also that a convex subset of a finite-dimensional Euclidean space has empty interior iff it is "flat", i.e., lies in some proper affine linear subspace. – Pete L. Clark Feb 28 '13 at 22:15
  • What is $[xz]$? How do we know that $[xz]\subset A$? It looks like the "closed line segment" except that later you use a comma for a half-open interval. – Thomas Andrews Mar 01 '13 at 03:11
  • @ThomasAndrews A typo. It is closed line segment, while $(xy]$ is half-open line segment. Thanks, fixed. – dtldarek Mar 01 '13 at 07:49
  • I guess the claim is also right when phrased with "relative interior" instead of "interior"? – Dirk Mar 01 '13 at 08:07
  • @Dirk You need some minor tweaks, e.g. the ball in the proof has to have correct dimension, e.g. $B \cap \mathrm{relint}(A)$. – dtldarek Mar 01 '13 at 08:43