Get the set of every direction of S

**Lolyta** · October 21st 2011, 08:13 AM

Well, d is a direction of a set S if for all $x \in{S}, \lambda \geq{0}, x+\lambda d \in{S}$ .
Let $S=\left\{{(x_1,x_2): x_1-2x_2 \geq{-6}, x_1-x_2\geq{-2}, x_1\geq{0}, x_2 \geq{1}}\right\}$ I have to get the set of every direction of S.

And I have the answer but I don't know how to reach it. The answer is: d has two components $d=(d_1,d_2), d_1 = 2\alpha + \beta; d_2=\alpha: \alpha, \beta \geq{0}$ . Could anyone help me?
Thanks.

**Deveno** · October 21st 2011, 02:40 PM

a direction is not a number. it could be a vector (like say, a unit vector).

your set S is a subset of the first quadrant. it is bounded below by $x_2 = 1$ , bounded on the left by $x_1 = 0$ and bounded above by

$x_2 = x_1 + 2$ and $x_2 = \dfrac{1}{2}x_1 + 3$ .

of these two lines, the second one has the lesser slope, and eventually is the more restrictive.

so for $(x_1,x_2) + \lambda(d_1,d_2) \in S$ , whenever $(x_1,x_2) \in S$ for all $\lambda > 0$ we have to have:

$d_1 \geq 0,\ d_2 \leq \dfrac{1}{2}d_1$

that is, $(d_1,d_2)$ must lie between the rays $(\alpha,0)$ and $(\alpha,\alpha/2), \alpha \geq 0$ .

**Lolyta** · October 23rd 2011, 07:46 AM

Originally Posted by Deveno

a direction is not a number. it could be a vector (like say, a unit vector).

your set S is a subset of the first quadrant. it is bounded below by $x_2 = 1$ , bounded on the left by $x_1 = 0$ and bounded above by

$x_2 = x_1 + 2$ and $x_2 = \dfrac{1}{2}x_1 + 3$ .

of these two lines, the second one has the lesser slope, and eventually is the more restrictive.

so for $(x_1,x_2) + \lambda(d_1,d_2) \in S$ , whenever $(x_1,x_2) \in S$ for all $\lambda > 0$ we have to have:

$d_1 \geq 0,\ d_2 \leq \dfrac{1}{2}d_1$

that is, $(d_1,d_2)$ must lie between the rays $(\alpha,0)$ and $(\alpha,\alpha/2), \alpha \geq 0$ .

Thanks for answering but I don't see how do you get those inequalities: $d_1 \geq 0,\ d_2 \leq \dfrac{1}{2}d_1$ .

Thank you again.

**Deveno** · October 23rd 2011, 08:24 AM

it might help if you draw a picture of S.

what happens if we add $(d_1,d_2)$ to the point (0,1) (which is in the lower left corner of S), if $d_1 < 0$ ? isn't the first coordinate going to shift to the left? that will put us outside of S. so we know we can only move to the right.

that leaves up-and-down. now, i should have included the condition $d_2 \geq 0$ as well, because if we go down, the second coordinate could drop below 1, and that will cause us to leave S as well.

so, we can only go up and to the right. well that's 3/4ths of the possible directions eliminated. you can think of this as all slopes between 0 and $\infty$ going up and to the right.

but now, if the slope of our ray $\lambda(d_1,d_2)$ is greater than 1/2, we'll eventually cross the line:

$x_2 = \frac{1}{2}x_1 + 3$ , which will take us outside of S. at best, we can only go parallel to it. this limits which directions we can go, for $(d_1,d_2)$ to work,

$d_2$ has to be less than or equal to 1/2 of $d_1$ .

it's a lot easier to understand all this, if you draw S, because then you can SEE it.

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