Let $X \subset \mathbb{R}^2$ be a set satisfying the following properties:
$(i)$ if $(x_1,y_1)$ and $(x_2,y_2)$ are any two distinct elements in $X$, then
either $x_1>x_2~\text{and}~y_1>y_2~~~~~~$ or $~~~~~~x_1<x_2~\text{and}~y_1<y_2$

$(ii)$ there are two elements $(a_1,b_1)$ and $(a_2,b_2)$ in $X$ such that for any $(x,y) \in X$,
$a_1\le x \le a_2 \text{ and } b_1\le y \le b_2$

$(iii)$ if $(x_1,y_1)$ and $(x_2,y_2)$ are two elements of $X$, then for all $\lambda \in [0,1]$,
$\left(\lambda x_1+(1-\lambda)x_2, \lambda y_1 + (1-\lambda)y_2\right) \in X$

Show that if $(x,y) \in X$, then for some $\lambda \in [0,1]$,
$x=\lambda a_1+(1-\lambda)a_2, y=\lambda b_1+(1-\lambda)b_2 $

The given conditions looked like convex function, but I could not use it to some good. Please help.


1 Answer 1


Fix elements $(a_1,b_1), (a_2,b_2)\in X$ such that (ii) holds with these.

Assume $(x,y)\in X$. Then by (ii), there exists a $\lambda\in[0,1]$ with $$\tag1x=\lambda a_1+(1-\lambda)a_2$$ (namely, if $a_1\ne a_2$, let $\lambda=\frac{a_2-x}{a_2-a_1}$; if $a_1=a_2$, any $\lambda$ will do). Apply (iii) to the points $(a_1,b_1)$ and $(a_2,b_2)$ to conclude that $(\lambda a_1+(1-\lambda)a_2,\lambda b_1+(1-\lambda)b_2)\in X$. If $(\lambda a_1+(1-\lambda)a_2,\lambda b_1+(1-\lambda)b_2)$ were distinct from $(x,y)$ then by (i) we'd have $\lambda a_1+(1-\lambda)a_2\ne x$. According to $(1)$, this is not the case, hence we conclude $$(x,y)=(\lambda a_1+(1-\lambda)a_2,\lambda b_1+(1-\lambda)b_2).$$

  • $\begingroup$ In this case, aren't we verifying by what is given, and not proving anything? Actually, this is what I had thought too...but I thought this was wrong...can you please tell me if I am wrong in thinking that? $\endgroup$
    – Hawk
    Commented May 4, 2014 at 13:32
  • $\begingroup$ Since my post ends in "hence we conclude ..." it looks a lot like proving something. Is there any specific step that makes you doubt? $\endgroup$ Commented May 4, 2014 at 14:31
  • $\begingroup$ yes, but please see this approach of mine...I think you will be able to understand it what I meant-math.stackexchange.com/questions/780834/… $\endgroup$
    – Hawk
    Commented May 4, 2014 at 14:57

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