1. The problem statement, all variables and given/known data 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? 2. Relevant 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. 3. 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 lets 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.