Is the statement 'If E is an open connected set then it is convex' true?

[EV33]

Homework Statement

It would make one of my proofs easy if it is true that

" If E is an open connected set then it is convex''.

I have been spending some time trying to prove this. Is this statement even true?

Homework Equations

Convex implies that if x is in E and y is in E then
εx+(1-ε)y is in E where 0<ε<1.

Connected: E cannot be the union the union of two nonempty separated sets.

The Attempt at a Solution

My attempt at the solution looks like this.

Suppose E is open in ℝⁿ. Since E is open implies that there exists a ball around every x in E s.t. b(x) is in E.

Let's choose a y in E s.t. b(y) intersected with b(x) is nonempty. Let's choose a point y* from the intersection. Let's also choose a x* in b(x). Since y* is also in b(x) and all balls are convex implies that εx*+(1-ε)y* is in b(x) union with b(y), where 0<ε<1

Now let's choose another point z s.t. b(z) intersected with b(y) is nonempty. Then similarly,
εy*+(1-ε)z* is in the union of b(x),b(y), and b(z).

Now assuming all this is correct... I am stuck here but I feel like if I can show
εx*+(1-ε)z* is in the union of b(x),b(y), and b(z) then the union of the 3 balls is convex which would imply that E is convex.

I also didn't mention connectedness but I think that comes into play as follows: I can find balls that have no separation and I can union them allowing me to create a path of balls connecting all points.
Thank you.

 

[chiro]

Hey EV33

This is just a suggestion based on the intuition of convex sets, but if a set were to be concave, then you could easily show an example of a separated set.

To get an idea of this with regards to the nature of the definition of a convex set, in a non-convex set (concave set) there will be a line that you can draw where one at least one part of that line does not lie in the boundary of the set.

Using this idea you can show that for a concave set, that you can form sets that are separated and thus don't have the property you have specified above.

 

[EV33]

Maybe I am interpreting what you are saying incorrectly but I think that you are talking about the converse of the "theorem" I stated. I want to show that if a set is open and connected then it is convex. I was not trying to say that if a set is convex then it must be connected.

 

[HallsofIvy]

Let A= {(x,y)| y= 0, $0\le x\le 1$}

Let B= {(x, y)| x= 0, $0\le y\le 1$}

Let C= {(x, y)| x= 1, $0\le y\le 1$}

Let D be the union of A, B, and C.

 

[Ray Vickson]

Homework Statement

It would make one of my proofs easy if it is true that

" If E is an open connected set then it is convex''.

I have been spending some time trying to prove this. Is this statement even true?

Homework Equations

Convex implies that if x is in E and y is in E then
εx+(1-ε)y is in E where 0<ε<1.

Connected: E cannot be the union the union of two nonempty separated sets.

The Attempt at a Solution

My attempt at the solution looks like this.

Suppose E is open in ℝⁿ. Since E is open implies that there exists a ball around every x in E s.t. b(x) is in E.

Let's choose a y in E s.t. b(y) intersected with b(x) is nonempty. Let's choose a point y* from the intersection. Let's also choose a x* in b(x). Since y* is also in b(x) and all balls are convex implies that εx*+(1-ε)y* is in b(x) union with b(y), where 0<ε<1

Now let's choose another point z s.t. b(z) intersected with b(y) is nonempty. Then similarly,
εy*+(1-ε)z* is in the union of b(x),b(y), and b(z).

Now assuming all this is correct... I am stuck here but I feel like if I can show
εx*+(1-ε)z* is in the union of b(x),b(y), and b(z) then the union of the 3 balls is convex which would imply that E is convex.

I also didn't mention connectedness but I think that comes into play as follows: I can find balls that have no separation and I can union them allowing me to create a path of balls connecting all points.



Thank you.

Suppose $A = \{(x,y) : x^2 + y^2 < 1\} \mbox{ and } B = \{(x,y) : (x-1)^2 + y^2 < 1 \}.$ These are open discs of radius 1 centered at (0,0) and (1,0). If C is their union, is C open? Is C connected? Is C convex?

RGV

 

[EV33]

If I have the two balls A and B and I union them then their union will be open and it will be connected. If I tried to draw a straight line connecting their tops I would fail.

So this implies that the theorem I was trying to prove is false right?


Thank you all for the help.

 

[HallsofIvy]

If I have the two balls A and B and I union them then their union will be open and it will be connected. If I tried to draw a straight line connecting their tops I would fail.

You mean, "will NOT be connected".

So this implies that the theorem I was trying to prove is false right?


Thank you all for the help.

Yes, your statement "an open connected set is convex" is false. The U shaped example I gave also shows that.