# to prove XY-plane a convex set??

we know that R and R^2 have convex subsets because any two points in them can be joined by a line segment..but how we will prove it mathematically that XY-plane is a convex set??

• From the definition of convexity? Take arbitrary $(x_1,y_1)$ and $(x_2,y_2)$ in $\mathbb{R}^2$... – newbie Feb 11 '14 at 14:27
• let x1=(1,0,0) and x2=(0,1,0). then by definition(1-alpha)x1+alpha x2 =x.now taking alpha=1/2 we get (1-1/2)(1,0,0)+1/2 (0,1,0)=(1/2,1/2,0) belongs to XY-plane...so XY-plane is a convex set.is this solution correct?? – hafsah Feb 11 '14 at 14:38
• the example is correct, but this is not a proof for "the XY-plane is convex". – Danny Feb 11 '14 at 14:43
• i want its prove .can u help me plz?? – hafsah Feb 11 '14 at 14:44
• Do you intuitively understand why the XY-plane is convex? – dani_s Feb 11 '14 at 14:49

After your comment, i think we talk about $\mathbb{R}^{3}$.
Therefore the XY-Plane is represented by the set of vectors $\left\{(x,y,0)^{T}|\;x,y\in\mathbb{R}\right\}$.
A set is convex, if for any vectors $\vec{a},\vec{b}$ the linesegment between $\vec{a}$ and $\vec{b}$ is in the set, too.
So the XY-plane is convex, if for all $\vec{a},\vec{b}\in\left\{(x,y,0)^{T}|\;x,y\in\mathbb{R}\right\}$ the following holds:

• $\lambda\vec{a}+(1-\lambda)\vec{b}\in\left\{(x,y,0)^{T}|\;x,y\in\mathbb{R}\right\},\;\;0\leq\lambda\leq1$

Let $\vec{a}=(x_{1},y_{1},0)^T$ and $\vec{b}=(x_{2},y_{2},0)^T$ be arbitrary vectors of the XY plane. Then

$\lambda\vec{a}+(1-\lambda)\vec{b}= \left( \begin{array}{c} \lambda x_{1}+(1-\lambda)x_{2}\\ \lambda y_{1}+(1-\lambda)y_{2}\\ 0\\ \end{array} \right)\in\left\{(x,y,0)^{T}|\;x,y\in\mathbb{R}\right\}$

because the $x$-component and the $y$-component are in $\mathbb{R}$ and the last component is $0$.

• if we want to talk about in R2 then we will not take the third copmonent and then prove it as above...is it so?? – hafsah Feb 11 '14 at 15:07
• If we talk about $\mathbb{R}^{2}$, then the XY-plane is the whole $\mathbb{R}^{2}$ and then the convexity is trivial. But yes, the proof works for $\mathbb{R}^{2}$ in the same way without the third coordinate.. – Danny Feb 11 '14 at 15:09