## convex sets

Hello! I've got some questions concerning convex sets.
We've had a lecture about convex sets this week, and got a some basic problems to solve. I think I can't use the material in the lecture to solve the problem. I'm just not sure about whether I fully understand the concept and can use it properly.
In the lecture the prof presented a proof for the fact that for every $t\in\matbb{R}$ the interval $I=[0,t]:=\{s|s\in\mathbb{R},0\le{s}\le{t}\}$ is convex.
The proof went like this.
Let $s_1,s_2\in[0,t]$ and $\lambda\in[0,1]$. The following holds: $0\le{s_1}$ and $s_2\le{t}$.
$\lambda{s_1}+(1-\lambda){s_2}\le\lambda{t}+(1-\lambda)t=t$. Hence, the interval is convex.
Now, the first question.
Why cannot we simply check whether the condition holds for $\lambda=0$ or $\lambda=1$ ?
It's $\lambda\cdot{0}+(1-\lambda)t$. Then for $\lambda=0$ t is in I and for $\lambda=1$ 0 is in I.
It's now asked to show that a hyperplane $E\subseteq\mathbb{R}^n$, $E:=\{(x_1,...,x_n)\in\mathbb{R}^n|a_1{x_1}+...+a_n{x_n}=b\}$, where $a_1,...,a_n,b\in\mathbb{R}$ and $(a_1,...,a_n)\neq(0,...,0)$ is a convex set.
Here, equally, it seems too easy just to put in two elements, say $x_1$ and $x_n$ and let $\lambda$ be either 0 or 1. If it's wrong to take this way, how can I show that E is convex differently?
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 Quote by symplectic_manifold Hello! I've got some questions concerning convex sets. We've had a lecture about convex sets this week, and got a some basic problems to solve. I think I can't use the material in the lecture to solve the problem. I'm just not sure about whether I fully understand the concept and can use it properly. In the lecture the prof presented a proof for the fact that for every $t\in\matbb{R}$ the interval $I=[0,t]:=\{s|s\in\mathbb{R},0\le{s}\le{t}\}$ is convex. The proof went like this. Let $s_1,s_2\in[0,t]$ and $\lambda\in[0,1]$. The following holds: $0\le{s_1}$ and $s_2\le{t}$. $\lambda{s_1}+(1-\lambda){s_2}\le\lambda{t}+(1-\lambda)t=t$. Hence, the interval is convex. Now, the first question. Why cannot we simply check whether the condition holds for $\lambda=0$ or $\lambda=1$ ? It's $\lambda\cdot{0}+(1-\lambda)t$. Then for $\lambda=0$ t is in I and for $\lambda=1$ 0 is in I.
You are given that the two points corresponding to$\lambda= 0$ and $\lambda= 1$ are in the set! The thing you want to prove is that every point on the interval between them is in the set- that corresponds to $0< \lamba< 1$.
Suppose you were asked to prove that the set [0, 1)union(1,2] was not convex. What would you do?

 It's now asked to show that a hyperplane $E\subseteq\mathbb{R}^n$, $E:=\{(x_1,...,x_n)\in\mathbb{R}^n|a_1{x_1}+...+a_n{x_n}=b\}$, where $a_1,...,a_n,b\in\mathbb{R}$ and $(a_1,...,a_n)\neq(0,...,0)$ is a convex set. Here, equally, it seems too easy just to put in two elements, say $x_1$ and $x_n$ and let $\lambda$ be either 0 or 1. If it's wrong to take this way, how can I show that E is convex differently?
First, x1 and x2 are not "elements" of Rn. They are simply components of a single point. You would have to take two arbitrary points, say (x1,x,...,xn) and (y1, y2, ... , yn) in the set and show that any point on the line segment between them (which can be written (x1+ $\lambda$(y1-x1), x2+ $\lambda$(y2-x2), ..., xn+ $\lambda$(yn-n1)) for $0< \lambda< 1$. $\lamba$= 0 or 1 just gives the points you started with. By hypothesis, they are in the set.
 Thanks for your reply! OK, for the union $I=[0,1)\cup(1,2]$ is it right if I do it this way?: Let $s_1,s_2\in{I}$ be two elements, such that $0\le{s_1}<1$ and [latex]1