TIFR CSE 2017 | Part A | Question: 13

Question

A set of points $S \subseteq \mathbb{R}^2$ is convex if for any points $x, \: y \: \in S$, every point on the straight line joining $x$ and $y$ is also in $S$. For two sets of points $S, T \subset \mathbb{R}^2$, define the sum $S+T$ as the set of points obtained by adding a point in $S$ to a point $T$. That is, $S+T := \{(x_1, x_2) \in \mathbb{R}^2 : x_1=y_1+z_1, \: x_2 = y_2 + z_2, \quad (y_1, y_2) \in S, (z_1, z_2) \in T\}$. Similarly, $S-T:=\{(x_1, x_2) \in \mathbb{R}^2 : x_1 = y_1-z_1, \: x_2=y_2-z_2, \quad (y_1, y_2) \in S, (z_1, z_2) \in T \}$ is the set of points obtained by subtracting a point in $T$ from a point in $S$. Which of the following statements is TRUE for all convex sets $S, \: T$?

• A. $S+T$ is convex but not $S-T$
• B. $S-T$ is convex but not $S+T$
• C. exactle one of $S+T$ and $S-T$ is convex, but it depends on $S$ and $T$ which one
• D. neither $S+T$ nor $S-T$ is convex
• E. both $S+T$ and $S-T$ are convex

Answer 1

Answer should be (E) Both will fill the space accordingly :
For ex :
                  
Now since the points are set therefore the closure property would hold on union and intersection because the space will move accordingly.
Note : Take a bunch of points as set and try to perform the operation b/w those sets all points will come in linear straight line therefore it will be convex too.

Comments

Can you please give an example?

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Set A ={(1,1)(2,2) (3,3)}
Set B={(0,3)(1,4)(2,5)}
SET (A+B) for one point i am deriving for ex : for (1,4) in set B ,Set(A+B) will contain {(2,5),(3,6)(4,7)}which is convex too .
PS : I took very trivial example but can be extended for all complex examples by taking general approach.

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Thank you very much