Convex Sets - Interior and Closure

**utopiaNow** · October 1st 2009, 09:36 PM

A set $E$ in $R^n$ is called convex if for every pair of points $x$ and $y$ in $E$ and every real $\lambda \in (0, 1)$ ,we have $[\lambda x + (1- \lambda)y] \in E$

1. Prove that the interior of convex sets is convex.
My attempt: Let $S^{\circ}$ be the set of interior points of $S$ . Choose $x \in S^{\circ}$ , this means that $\exists r > 0$ such that $B(x;r) \subseteq S$ . Now choose two points in this open ball, $y$ and $z,$ since $y, z \in B(x;r)$ they are also interior points of $S$ . And since every open ball is convex, $\lambda y + (1-\lambda)z \in B(x;r)$ and thus $\lambda y + (1-\lambda)z \in S^{\circ}$ , as required.

I have a problem with that "proof" since I never used the convexity of S, I basically proved that any interior set is convex. Or that every open set is convex. But I can't find where I'm going wrong, or a different proof that makes the use of the convexity of S.

2. Prove that the closure of a convex set is convex.
My attempt, What I have so far: Let $\overline{E}$ be the closure of $E$ . Pick arbitrary $x, y \in \overline{E}$ , I need to show $\lambda x + (1- \lambda)y \in \overline{E}$ .
3 cases:
(i) $x,y \in E$ ; I showed the implication from here very easily.
(ii) $x,y \in E'$ , where $E'$ is the set of all the limit points of $E$ .
(iii) Without loss of generality, $x \in E$ and $y \in E'$ .

I'm having problem with the last "2" cases as I don't know how to proceed from the fact that something is a limit point to showing that $\lambda x + (1- \lambda)y \in \overline{E}$ .

Any suggestions would be appreciated. Thanks, in advance.

**InvisibleMan** · October 2nd 2009, 04:03 AM

Originally Posted by utopiaNow

1. Prove that the interior of convex sets is convex.
My attempt: Let $S^{\circ}$ be the set of interior points of $S$ . Choose $x \in S^{\circ}$ , this means that $\exists r > 0$ such that $B(x;r) \subseteq S$ . Now choose two points in this open ball, $y$ and $z,$ since $y, z \in B(x;r)$ they are also interior points of $S$ . And since every open ball is convex, $\lambda y + (1-\lambda)z \in B(x;r)$ and thus $\lambda y + (1-\lambda)z \in S^{\circ}$ , as required.

Yes, you are right there's a problem, you need to show that for every y,z in Int(S) the line segment joining them is also in Int(S).
There's a famous proof that does it, you first pick a disk around y, if z is contained in this disk then you are done, if not then the point ay+(1-a)z is in this disk (for some a in [0,1]), not there's another disk with centre around this point, if z is contained in this disk then we are done if not procceed as in the preceding step, this problem should halt in a finite number of steps with the last point being z and so the line joining y and z is in the interior of S as well.

2. Prove that the closure of a convex set is convex.
My attempt, What I have so far: Let $\overline{E}$ be the closure of $E$ . Pick arbitrary $x, y \in \overline{E}$ , I need to show $\lambda x + (1- \lambda)y \in \overline{E}$ .
3 cases:
(i) $x,y \in E$ ; I showed the implication from here very easily.
(ii) $x,y \in E'$ , where $E'$ is the set of all the limit points of $E$ .
(iii) Without loss of generality, $x \in E$ and $y \in E'$ .

I'm having problem with the last "2" cases as I don't know how to proceed from the fact that something is a limit point to showing that $\lambda x + (1- \lambda)y \in \overline{E}$ .

Any suggestions would be appreciated. Thanks, in advance.

For the limit points, you can use the fact that there is a sequence of points in E which converge to the limit point, for example suppose:
x is a limit point and y is a limit point, then for each of them, there are sequence points converging to them: {x_n}, {y_n}, so the second problem is translated to the third problem of proving that if one is a limit point and the other is interior point, in this case just look at
$\lambda x +(1-\lambda)y_n$ where y_n are interior points then for each n these line segment are in E, now if the limit of this sequence of lines was crossing cl(E) with its exterior then at least one of the other lines in the sequence of lines will cross cl(E) with its exterior, thus not contained as a whole in cl(E) opposite to our assumption.

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