# Thread: Prove a set is convex algebraically

1. ## Prove a set is convex algebraically

How can I determine if a set is convex without graphing it?

I have that a set, $\displaystyle S$, is convex if, given any two points, $\displaystyle x_1, x_2 \in S$, the line connecting them is entirely within $\displaystyle S$, or:

$\displaystyle \lambda x_1 + (1-\lambda) x_2 \in S \qquad \forall \lambda \in [0,1]$

Some of the sets in question:
b. $\displaystyle \{(x,y,z) \colon z=\left| y \right|, x \le 3 \}$
c. $\displaystyle \{(x,y): y-3x^2=0\}$
d. $\displaystyle \{(x,y,z): y \ge x^2, x+2y+z \le 4\}$
e. $\displaystyle \{(x,y):x=3, \left| y \right|, \le 4 \}$
f. $\displaystyle \{(x,y,z): z= \left| y \right|, x \le 3 \}$

I have to hand in f. I'm not good enough to draw it, but I can envision it. It looks like a "V" going from $\displaystyle x=3$ down to $\displaystyle -\infty$. Clearly, it is not convex. But how would I show that mathematically?

Thanks.

2. Originally Posted by MSUMathStdnt
Some of the sets in question (I can't make the braces in math mode?):

If you write \{a\}, you'll obtain $\displaystyle \{a\}$ .

b. {$\displaystyle (x,y,z) \colon z=\left| y \right|, x \le 3$}

It is not convex. Choose:

$\displaystyle x_1=(0,-1,1),\;x_2=(0,1,1),\;\lambda=1/2$.

Fernando Revilla