Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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
1  
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
show 2 more comments

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.