Convex sets in the plane

dickchimney · Poincare Conjecture Offline Joined: 06 Mar 2004 Posts: 218

Given $A$ a set of points in the plane.
Assume that $A$ is not convex.
Let $B$ be the union of the segments whose endpoints belong to $A$ , namely, if $M,N$ are in $A$
then all the points of the segment $MN$ are in $B$ . Prove or disprove $B$ is convex provided that
a) $A$ is path-connected
b) $A$ is connected
I would be very appreciate if anyone can devise any weaker condition Very Happy

**Posted:** Wed Jan 05, 2005 2:48 am

Myth

What is connected set in your opinion?

**Posted:** Wed Jan 05, 2005 11:17 am

_________________
Myth is out of here

grobber

I think the answer is positive in the first case Confused

Take any point $P$ belonging to the convex hull of $A$ . This point must then be the convex linear combination of a few points in $A$ , meaning that it lies inside a convex polygon $A_1A_2\ldots A_n$ , with vertices among the points of $A$ . For each $i\in\overline{1,n}$ , let $\mathcal L_i$ be a continuous path connecting $A_i,A_{i+1}$ (the indices are $\pmod n$ ).

Now consider the case when there is an $i$ s.t. there is a homotopy $H_i(x,t)$ taking $\mathcal L_i$ to $A_iA_{i+1}$ with the property that $H_i(x,t)=P$ , for some $x,t$ , and the case when there are no such homotopies for any $i$ .

The results seem very intuitive, but I think the proofs should not be that easy, since we deal with continuous curves, and we know what a pain these can be (think about the Jordan Curve Theorem).

**Posted:** Wed Jan 05, 2005 11:25 am

grobber

To Myth: I understood it to be the usual definition of connectedness: there are no two sets $A_1,A_2$ s.t. $A_1\cup A_2=A$ and $\overline{A_1}\cap A_2=\overline{A_2}\cap A_1=\emptyset$ . Or, equivalently, there's no proper subset of $A$ which is both open and closed in the topology induced by the usual topology on $\mathbb R^2$ .

Maybe dickchimeny means something else?

**Posted:** Wed Jan 05, 2005 11:32 am

dickchimney · Poincare Conjecture Offline Joined: 06 Mar 2004 Posts: 218

**Posted:** Wed Jan 05, 2005 9:36 pm

grobber

I didn’t manage to make it very concise, but I’ll try.

We have to show that each point in the convex hull belongs to the set $B$ . Take such a point $P$ . Since $P$ is in the convex hull, we can interpret this by saying that no matter which line through $P$ we choose, there are points of the set $A$ in both half-planes determined by the line (disregard the otherwise easy to treat limit cases when $P$ lies on the boundary of the convex hull). On the other hand, since we assume that $P$ does not lie on a segment with endpoints in $A$ , given a line through $P$ , there is a point from $A$ on precisely one of the rays belonging to the line and starting from $P$ .

For simplicity, identify each ray starting from $P$ with $x\in[0,1)$ s.t. the ray has the direction of $e^{2\pi ix}$ .

If there are no points from $A$ on the rays $x$ and $y>x$ , then there must be no points from $A$ in one of the angles determined by the two rays, namely the smaller one (otherwise we could split the set $A$ in two sets, one inside each angle determined by the two rays, and we would contradict the connectedness). This will show fairly easily that there are, infact, no points from $A$ on the rays corresponding to $x\in$ an entire interval of length $\frac 12$ , closed at one end and open at the other one, and this will contradict the fact that $P$ is in the convex hull.

It’s a bit unclear, and I don’t think I have all the details worked out, but even if it’s flawed, it should give a starting point for further investigations Smile

.

**Posted:** Fri Jan 07, 2005 4:06 am

dickchimney · Poincare Conjecture Offline Joined: 06 Mar 2004 Posts: 218

Thanks for the proof!! But can you make it clearer??
For example, at each step you indicate what you're going to prove?
By this problem, I'm asking , not quizing! So feel free to explain it clearly for me Mr. Green

**Posted:** Fri Jan 07, 2005 11:58 am

grobber

I'll help, but which part should I explain better?

**Posted:** Fri Jan 07, 2005 6:13 pm

dickchimney · Poincare Conjecture Offline Joined: 06 Mar 2004 Posts: 218

**Posted:** Fri Jan 07, 2005 11:48 pm

grobber

Yes

.

I have to apologize about that. Before writing it, I had another message saying "I think b) works as well (blahblahblah)". After that, I erased my initial message, and posted this one, but forgot to mention what I was proving Smile

. Sorry again. Indeed, that message is concerned with b).

**Posted:** Sat Jan 08, 2005 12:16 am

dickchimney · Poincare Conjecture Offline Joined: 06 Mar 2004 Posts: 218

So correct!! Mr. Green

Thanks, infact b) is the part I need!! Smile

**Posted:** Sat Jan 08, 2005 10:42 am

grobber

b) is, of course, stronger than a), so we have the problem secured pretty good now Smile

. What weaker conditions did you have in mind?

**Posted:** Sat Jan 08, 2005 10:46 am

dickchimney · Poincare Conjecture Offline Joined: 06 Mar 2004 Posts: 218

I have no idea
But, as you said, the problem is pretty well-secured now. Wink

**Posted:** Mon Jan 10, 2005 9:22 pm