# Prove that if $(x,y)∈X$, then for some $λ\in [0,1],\space x=λa_1+(1−λ)a_2,\space y=λb_1+(1−λ)b_2$.

Let $$X⊂R^2$$ 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, $$x_1>x_2$$ and $$y_1>y_2$$ or, $$x_1 and $$y_1
(ii) There are two elements $$(a_1,b_1)$$ and $$(a_2,b_2)$$ in $$X$$ such that for any $$(x,y)\in X$$, $$a_1≤x\leq a_2$$ and $$b_1≤y≤b_2$$
(iii) If $$(x_1,y_1)$$ and $$(x_2,y_2)$$ are two elements of $$X$$, then for all $$λ∈[0,1],\space (λx_1+(1−λ)x_2,λy_1+(1−λ)y_2)∈X$$
Show that if $$(x,y)∈X$$, then for some $$λ\in [0,1],\space x=λa_1+(1−λ)a_2,\space y=λb_1+(1−λ)b_2$$

This question is the same as the one I am asking here but I think that my approach is different from the one given there and I want to verify my approach.

My Approach:

From observation $$(i)$$, it can be said that $$x_1-x_2$$ and $$y_1-y_2$$ will always be of same sign. Hence the slope of line passing through points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ will be positive.

From observation $$(ii)$$, which says that if $$(x,y)\in X$$, then $$a_1\leq x\leq a_2$$ and $$b_1\leq y\leq b_2$$, it can be inferred that the subset $$S$$ denotes a rectangle in $$R^2$$ with vertices $$(a_i,b_i),\space i,j=1$$ or $$2$$. (suppose this)

From observation $$(iii)$$, the point can be written as $$(\lambda(x_1-x_2)+x_2,\space \lambda(y_1-y_2)+y_2)$$. It can be observed that this point satisfies the equation $$y-y_2=\dfrac{y_1-y_2}{x_1-x_2}(x-x_2)$$. Thus any such point will lie on the line joining $$(x_1,y_1)$$ and $$(x_2,y_2)$$. Also since $$\lambda\in [0,1],\space x_2\leq (\lambda(x_1-x_2)+x_2)\leq x_1$$ and $$y_2\leq (\lambda(y_1-y_2)+y_2)\leq y_1$$ (assuming $$x_1>x_2$$ and $$y_1>y_2$$). Thus this point will always lie on line segment joining the points $$(x_1,y_1)$$ and $$(x_2,y_2)$$.

Using these observations, we can say that $$(\lambda(a_1-a_2)+a_2,\lambda(b_1-b_2)+b_2)$$ or $$(λa_1+(1−λ)a_2,λb_1+(1−λ)b_2)$$ will lie on line segment joining $$(a_1,b_1)$$ and $$(a_2,b_2)$$.

Since the given line segment lies completely in $$X$$, there must be some point $$(x,y)\in X$$ which corresponds to the given point.

After writing this complete solution, I have now observed that I have done the opposite of what was asked, but I feel that this approach of connecting this question to coordinate geometry must yield something good. So please offer suggestions to improve this solution but using the same approach.

THANKS

• "it can be inferred that the subset 𝑆 denotes a rectangle" Might be misreading this, but it seems you are claiming here that that the set has to be the whole rectangle? Condition (ii) only implies that the set is contained in a rectangle. Commented Sep 6, 2020 at 20:13
• So essentially you want to show that under the given conditions, $X$ is a diagonal of the rectangle. I haven't thought this through in detail, but assume it is not (draw a picture), and by drawing lines between known points, I think it'd be easy to find two points violating condition (i). Commented Sep 6, 2020 at 20:18
• @rogerl, I'm not sure how do I prove that $X$ is the diagonal of the rectangle using this approach. Please suggest something. Commented Sep 6, 2020 at 20:25
• If there is any points outside of this line in the set then condition (i) will be violated. To simplify it for yourself you can assume $(a_1,b_1) = (0,0)$ and $(a_2,b_2) = (1,1)$ so that all points $(t,t)$ with $t\in[0,1]$ are in $X$. Now if for example $(0.3, 0.4)$ was in $X$ then consider the point $(0.35, 0.35) \in X$. This point has $0.3 - 0.35 = -0.05$ and $0.4 - 0.35 = 0.05$ so condition (i) is violated. Thus one approach is to assume that we have a point outside the line in $X$ and then show that this leads to a contradiction. Commented Sep 6, 2020 at 20:31
• @Winther, thanks for the help. Will anyone of the commentators write it as an answer so that I may accept it? Commented Sep 6, 2020 at 20:33

Condition (ii) and (iii) implies that the line connecting $$(a_1,b_1)$$ and $$(a_2,b_2)$$ is in $$X$$. To prove that this line is the whole of $$X$$ you need to show that any point outside of this line would give a contradiction to Condition (i). To do this note that any point $$(x,y)$$ inside the rectangle defined by Condition (ii) can be written on the form $$x = a_1\lambda_1 + (1-\lambda_1)a_2$$ and $$y = b_1\lambda_2 + (1-\lambda_2)b_2$$ for some $$\lambda_1,\lambda_2\in [0,1]$$ and the problems asks you to prove that $$\lambda_1 \equiv \lambda_2$$. Compare the point $$(x,y)$$ to the point $$(x^*,y^*)$$ defined the same way as above but with $$\lambda = \frac{\lambda_1+\lambda_2}{2}$$ for both coordinates and check Condition (i).
You can also do this purely geometrically: given a point outside the line draw a normal vector from the line to the point and consider the components of this vector. Can both have the same sign? For this its useful to note that the vector from $$(a_1,b_1)$$ to the point on the line (which has positive components) dotted with the normal vector has to be zero.
Hint : Prove that the given conditions are equivalent to saying that $$X$$ is the graph in $$\mathbb{R}^2$$ of an affine (and strictly increasing) function defined on a segment of $$\mathbb{R}$$.