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Prove or disprove the following statement: A set X is convex if $\forall x,y \in X$, $\frac{1}{2}(x+y)\in X$.

A set X is convex is $\forall x,y\in X, \lambda \in [0,1] \implies \lambda x + (1-\lambda)y \in X$.
I think that the statement doesn't hold and I tried to disprove it formally, but I couldn't. Should I be looking for counterexamples? Any hints?

Martin Sleziak
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  • Let $X$ be a field of characteristic not equal to 2. – nobody Oct 14 '18 at 08:57
  • @Aritra That looks like a rhetorical statement. –  Oct 14 '18 at 08:57
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    It is true for closed sets: https://math.stackexchange.com/questions/1371147/prove-if-c-is-midpoint-convex-and-closed-then-its-a-convex-set. – Martin R Oct 14 '18 at 09:04
  • @Aritra What do you mean? Take $\mathbb Z_3$, what is $\lambda \cdot 2$ to you when $\lambda$ is a real number? – Git Gud Oct 14 '18 at 09:45
  • It is false in general as Saucy O'Path has pointed out below. However, if $X$ is a subset of a Hilbert space and $X$ is closed + your condition. Then $X$ is indeed convex. :) – weirdo Oct 17 '18 at 02:48

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That looks very much like a condition that $\Bbb Q$ would satisfy.

  • +1... Even if we restrict $\lambda$ to members of $\Bbb Q\cap [0,1],$ if $X={a2^{-b}:a\in \Bbb Z\land b\in \Bbb N} $ then $\forall x,y\in X;((x+y)/2\in X)$ But $0,1\in X$ while $(2/3)0+(1/3)1=1/3\not \in X.$ – DanielWainfleet Oct 14 '18 at 09:18