2

Let $S$ be the set $$S= \{(x,y) \in \mathbb R^2\mid x>1,y>1\}$$

Is this set convex and if so, how do you prove it? I've tried using the definition by looking at two arbitrary vectors in the set and looking at their convex combination, but have not been succesful thus far.

air
  • 2,812
  • 3
    You should show what you've tried. – Tim Raczkowski Sep 28 '15 at 17:31
  • I've let z=(x_1,y_1) and w=(x_2,y_2) and looked at tz+(1-t)w and gotten the equations

    (1)tx_1+(1-t)x_2= t(x_1-x_2)+x_2

    ty_1+(1-t)y_2= = t(y_1-y_2)+y_2

    Due to symmetry of x and y I only need to show that one of the equations are greater than 1 for all t \in [0,1]. First I looked at t=1 and then t=0, and concluded that its greater than 1 if this is the case.Then I assumed 0< t < 1 and x_1 > x_2 and it worked out and the same if I say 0<t<1 and x_1=x_2. The only case im having trouble with is t<0<1 and x_1<x_2 Appreciate the help! :)

     –  Sep 28 '15 at 17:57

3 Answers3

3

It is straightforward to show that if $C_1, C_2$ are convex sets, then so is $C_1 \cap C_2$ (this holds in general for any collection of convex sets).

Also, if $S_1,S_2$ are convex, then so is $S_1 \times S_2$.

Let $C_1 = \{ ((x,y) | x>1 \} = (1,\infty) \times \mathbb{R}$, $C_2 = \{ ((x,y) | y>1 \} = \mathbb{R} \times (1,\infty)$, and note that $S = C_1 \cap C_2$.

copper.hat
  • 172,524
3

Hint: the (non-empty) intersection of convex sets is convex, and the sets $\{(x,y):x>1\}$ and $\{(x,y):y>1\}$ are quite trivially convex (and not disjoint).

Jack D'Aurizio
  • 353,855
0

Hint: Let $v,w$ be two such vectors and $tv+(1-t)w$ be a convex combination. Can you show that the $x$ and $y$ components of this convex combination satisfy $x>1$ and $y>1$?

parsiad
  • 25,154