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As the title says, i want to show that a closed set $M \subset \mathbb{R^{n}}$ with $\forall x,y \in M, \frac{1}{2}(x+y) \in M$ is a convex set.

My idea was to show this by contraposition, so showing

$M$ is not convex $\implies$ $\frac{1}{2}(x+y) \notin M$.

By definition of a convex set and the negation of the statement:

$\forall t \in [0,1] \forall x,y \in M:tx+(1-t)y \in M$

We get:

$\exists t \in [0,1]$ with $tx+(1-t)y \notin M$

Now if we choose $t=\frac{1}{2}$, it follows that $\frac{1}{2}(x+y) \notin M$ and by contraposition the initial statement is proven.

I am not so sure if it is okay to just choose a $t$ here.

J3ck_Budl7y
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2 Answers2

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You know there exists a $t$, you cannot choose it.

Take $x,y\in M$. By the property that $M$ satisfies, one can prove by induction that $(1-k/2^n)x+k/2^ny\in M$ for every $n\in\mathbb{N}$ and $k=0,1,\ldots, 2^n$ (this set is obtained recursively finding all midpoints).

The set of all this points is dense in the segment $[x,y]$, that is, given any point $p$ in the segment $[x,y]$ we can find a point of that form (and thus, a point in $M$) arbitrarily close to $p$.

If it was the case that $p\notin M$, using that $M$ is closed, we could find a neighbourhood of $p$ with no elements of $M$. But that contradicts the density of the above set of points, so necessarily $p\in M$. As the segment between any two points in $M$ is contained in $M$, we conclude $M$ is convex, as desired.

Julio Puerta
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1

The statement

$\exists t\in[0,1]$ with $tx+(1−t)y\not\in M$

is not equivalent to

$\frac{1}{2} (x+y)\not\in M$

as the existence of a $t \in [0, 1]$ does not guarantee that $t = \frac{1}{2}$.

Idea of a solution

This exercise can be solved with a similar approach to the bisection method.

Proof

If $t \in \{0, 1\}$, the equality holds.

Let $t \in (0, 1)$ be a dyadic rational, i. e. $\exists\, m \in \mathbb{N}, \exists\, a \in [\![0, 2^m - 1]\!], t = \frac{a}{2^m}$ and $a \wedge 2^m = 1$ ($a$ is odd).

Let $a = \overline{a_1 a_2 ... a_{m-1} a_m}$ be the binary representation of $a$, with $a_m = 1$. Then $t = \overline{0.a_1 a_2 ... a_{m-1} a_m}$.

Let us show by induction on $m$ that $\forall\, x, y \in M, t x + (1 - t) y \in M$.

  • If $m = 1$, then $t = \overline{0.a_1} = \overline{0.1} = \frac{1}{2} = 1 - t$. By hypothesis, $tx + (1-t) y = \frac{1}{2}(x + y) \in M$.
  • Suppose that $\forall\, t \in [0, 1), (\exists\, a \in [\![0, 2^m - 1]\!], t = \frac{a}{2^m} \implies \forall\, x, y \in M, t x + (1 - t) y \in M)$.
    We want to show that property for $m + 1$.
    Let us write $u := \overline{0.a_2 a_3... a_m a_{m + 1}}$. By hypothesis of induction, $\forall\, x, y \in M, ux + (1 - u)y \in M$.
    • If $a_1 = 0$, then $\frac{1}{2} (\underbrace{ux + (1 - u)y}_{\in M} + y) \in M$.
      As $\frac{1}{2} (ux + (1 - u)y + y) = tx + (1-t)y \in M$, this concludes this case.
    • Similarly, if $a_1 = 1$, then $\frac{1}{2} (x + \underbrace{ux + (1 - u)y}_{\in M}) \in M$.
      As $\frac{1}{2} (x + ux + (1 - u)y) = tx + (1-t)y$, we get $tx + (1-t)y \in M$.

Conclusion: If $t$ is a dyadic rational, then $\forall\, x, y \in M, t x + (1 - t) y \in M$.

You can write $t = \overline{0.t_1 t_2 ... t_k...}$ the dyadic representation of t (in base 2), s. t. $$ t = \sum_{k = 1}^{+\infty} \frac{t_k}{2^k} $$

Writing $(v_m)_{m \in \mathbb{N}}$ the sequence of dyadic rationals $v_m := \overline{0.t_1 t_2 ... t_m}$, it follows that $$\forall\, x, y \in M, v_m x + (1- v_m) y \in M$$

As $M$ is closed and $v_m \to t$, we get $$\forall\, x, y \in M, t x + (1 - t) y \in M$$

Finally, $M$ is a convex set.

Jean
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