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.

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    $\begingroup$ Is this true? A line in $\mathbb{R}^2$ is convex, but has empty interior. $\endgroup$ – dtldarek Feb 28 '13 at 21:00
  • $\begingroup$ ok, I changed $\mathbb{R}^n$ to $\mathbb{R}$ $\endgroup$ – Ashot Feb 28 '13 at 21:02
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    $\begingroup$ Well, a single point is convex in $\mathbb{R}$... $\endgroup$ – dtldarek Feb 28 '13 at 21:04
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    $\begingroup$ In $\mathbb R$, the convex sets are very easily characterized. Per the above comment, you have to insist that $A$ have at least two points. For $\mathbb R^n$ you need to insist that there are $n+1$ points which are not all in the same $n-1$-plane. $\endgroup$ – Thomas Andrews Feb 28 '13 at 21:05
  • $\begingroup$ Related: math.stackexchange.com/questions/1671444 $\endgroup$ – Watson Nov 6 '16 at 20:03

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 ;-)

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    $\begingroup$ 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. $\endgroup$ – Pete L. Clark Feb 28 '13 at 22:15
  • $\begingroup$ 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. $\endgroup$ – Thomas Andrews Mar 1 '13 at 3:11
  • $\begingroup$ @ThomasAndrews A typo. It is closed line segment, while $(xy]$ is half-open line segment. Thanks, fixed. $\endgroup$ – dtldarek Mar 1 '13 at 7:49
  • $\begingroup$ I guess the claim is also right when phrased with "relative interior" instead of "interior"? $\endgroup$ – Dirk Mar 1 '13 at 8:07
  • $\begingroup$ @Dirk You need some minor tweaks, e.g. the ball in the proof has to have correct dimension, e.g. $B \cap \mathrm{relint}(A)$. $\endgroup$ – dtldarek Mar 1 '13 at 8:43

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