# Is $S^\circ$ convex if $S$ convex?

Suppose $S\subset \mathbb{R^n}$ and $S^\circ$ denoted as the interior of $S$.Is $S^\circ$ convex if $S$ convex? $S$ is Convex mean $\forall x,y\in S, kx+(1-k)y\in S, k\in [0,1]$ I know how to prove that $\bar S$ convex. For $S^\circ$ it seems obviously, i originally wanna claim that as $S^\circ\subset S$ so it must be convex. But i want to know if it is possible to use the definition of convexity of a set to prove that $S^\circ$ is convex.

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The reason you gave is not valid, because there are subsets of convex sets that are not convex. What is $S$? A subset of $\mathbb R^n$ or some other normed space? A subset of an arbitrary topological vector space? – Jonas Meyer Dec 5 '12 at 4:45
Try to prove that for every $k\in [0,1]$, and for all $x,y\in S^\circ$, it is $kx+(1-k)y\in S^\circ$. Hint: If $x\in S^\circ$, then there is an open neighborhood of $x$ in $S$, i.e. $x\in V$, $V$ is open, $V\subseteq S$. The same holds for $y$... – Pantelis Sopasakis Dec 5 '12 at 4:52
am sorry, but what does $S^0$ denote? – dineshdileep Dec 5 '12 at 5:07
You should draw a picture. The picture is (almost) the proof. – Olivier Bégassat Dec 5 '12 at 5:09
@JonasMeyer Do you mean that the argument $S^0\subset S$ so if $S$ is convex then $S^0$ is convex is false? – Mathematics Dec 5 '12 at 5:25

Let $X$ be a topological vector space over $\mathbb{R}$.
Recall the definition of open neighborhoods: $U$ is said to be an open neighborhood of $x\in X$ if it is an open set and it contains $x$.
First, $x\in S^\circ$ iff there is an open neighborhood $V_x\ni x$ such that $V_x\subseteq S$. Take $x,y\in S^\circ$ and $V_x,$ $V_y$ two open neighborhoods of $x$ and $y$ in $S$ respectively. We need to prove that $\lambda x + (1-\lambda)y\in S^\circ$ for all $\lambda\in (0,1)$. Since, $S$ is convex we have that $\lambda x + (1-\lambda)y\in S$ for all $\lambda\in (0,1)$. It suffices to find an open neighborhood of the point $\lambda x + (1-\lambda)y$ which lies inside $S$. This is:
$$V^\lambda = \lambda V_x + (1-\lambda)V_y = \{z=\lambda v_x + (1-\lambda)v_y;\ v_x\in V_x, v_y\in V_y\}$$
Since the mappings $+:X\times X\to X$ and $\cdot:\mathbb{R}\times X\to X$ are open mappings, $V^\lambda$ is open. Additionally, $V^\lambda\subseteq S$. Indeed, every $z\in V^\lambda$ is written as a convex combination of points of $S$. We have proved that $V^\lambda$ is an open neighborhood of $\lambda x + (1-\lambda)y$. This completes the proof.